Coordinate Geometry

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CLASS X Mathematics ~4 marks/year Ch 7 of 14
Coordinate Geometry

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Every point on a plane is pinned down by an ordered pair $(x,y)$ — the abscissa $x$ (distance from the $y$-axis) and the ordinate $y$ (distance from the $x$-axis).
  • The distance formula $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$ comes straight from Pythagoras — it gives the length of the segment joining two points.
  • The section formula finds the point that divides a segment in a given ratio $m_1:m_2$; the midpoint is just the special case $1:1$.
  • These two tools let us check shapes (triangle, square, parallelogram), prove points collinear, find missing coordinates and unknown ratios.
  • Board weightage: ~4 marks/year — typically one distance-formula question (find distance / prove a shape, 2–3 marks) and one section-formula question (ratio or midpoint, 2–3 marks).
Detailed notes

1. Where this chapter sits

In Class 9 you learnt how to locate a point on the plane using a pair of perpendicular axes. Quick recap:

  • The horizontal line is the $x$-axis, the vertical line is the $y$-axis, and they cross at the origin $O(0,0)$.
  • The distance of a point from the $y$-axis is its $x$-coordinate or abscissa; the distance from the $x$-axis is its $y$-coordinate or ordinate.
  • A point on the $x$-axis has the form $(x,0)$; a point on the $y$-axis has the form $(0,y)$.

You also saw that a linear equation $ax+by+c=0$ graphs as a straight line, and $y=ax^{2}+bx+c$ (with $a\neq0$) graphs as a parabola. Coordinate geometry is the bridge that lets us study geometry using algebra — measuring lengths, dividing segments and testing shapes purely by calculation. This chapter answers two questions: how far apart are two points, and where exactly does a point sit if it splits a segment in a given ratio?

The two axes cut the plane into four quadrants, numbered anticlockwise starting from the top right: Quadrant I has signs $(+,+)$, Quadrant II is $(-,+)$, Quadrant III is $(-,-)$ and Quadrant IV is $(+,-)$. Knowing the quadrant of a point is a quick sanity check on your answers — if a calculation places a point in the wrong quadrant, you have probably made a sign error. Remember too that the ordered pair matters: the point $(3,5)$ is completely different from $(5,3)$, because the first number is always read along the $x$-axis and the second along the $y$-axis. Coordinate geometry is widely applied in physics, engineering, navigation, seismology and computer graphics, so the small toolkit you build here reappears far beyond the exam hall.

2. Building the distance formula from Pythagoras

Start simple. If two points lie on the $x$-axis, say $A(4,0)$ and $B(6,0)$, the distance is just $OB-OA=6-4=2$ units. Likewise two points on the $y$-axis, $C(0,3)$ and $D(0,8)$, are $8-3=5$ units apart.

For a point off the axes we drop perpendiculars and use the right triangle. For $A(4,0)$ and $C(0,3)$: with $OA=4$ and $OC=3$, $AC=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$ units.

Now the general case. Take any two points $P(x_1,y_1)$ and $Q(x_2,y_2)$. Drop perpendiculars to the $x$-axis and complete a right triangle $PTQ$ with the right angle at $T$. Then the horizontal leg is $PT=x_2-x_1$ and the vertical leg is $QT=y_2-y_1$. By Pythagoras:

$$PQ^{2}=PT^{2}+QT^{2}=(x_2-x_1)^{2}+(y_2-y_1)^{2}$$

Distance is never negative, so we take the positive square root.

3. The distance formula

The distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ is:

$$PQ=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$$

Two useful remarks:

  • The distance of a point $P(x,y)$ from the origin $O(0,0)$ is $OP=\sqrt{x^{2}+y^{2}}$.
  • Order does not matter: $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}=\sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$, since squaring removes the sign.

Worked motivation (Fig. 7.3 / 7.4 in NCERT): for $P(4,6)$ and $Q(6,8)$ in the first quadrant, $PQ^{2}=2^{2}+2^{2}=8$, so $PQ=2\sqrt2$. For points in different quadrants like $P(6,4)$ and $Q(-5,-3)$, $PQ=\sqrt{11^{2}+7^{2}}=\sqrt{170}$. The same formula works everywhere.

4. Using the distance formula — shapes and collinearity

Most distance questions are really shape questions. The strategy: compute the side (and sometimes diagonal) lengths, then read off the shape.

  • Triangle: three points form a triangle if the sum of any two distances is greater than the third (they are not collinear).
  • Collinear: points $A,B,C$ are collinear if the largest distance equals the sum of the other two (e.g. $AB+BC=AC$).
  • Isosceles triangle: exactly two sides equal. Equilateral: all three equal.
  • Right triangle: the longest side squared equals the sum of the squares of the other two (converse of Pythagoras).
  • Square: all four sides equal and both diagonals equal. Rhombus: all four sides equal (diagonals need not be). Rectangle: opposite sides equal and diagonals equal. Parallelogram: opposite sides equal.
NCERT Example 1 — do $(3,2),(-2,-3),(2,3)$ form a triangle?

Let $P(3,2),\,Q(-2,-3),\,R(2,3)$. Then

$PQ=\sqrt{(3+2)^{2}+(2+3)^{2}}=\sqrt{25+25}=\sqrt{50}\approx7.07$,

$QR=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{16+36}=\sqrt{52}\approx7.21$,

$PR=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2}\approx1.41$.

The sum of any two of these exceeds the third, so $P,Q,R$ form a triangle. Also $PQ^{2}+PR^{2}=50+2=52=QR^{2}$, so by the converse of Pythagoras $\angle P=90^{\circ}$ — it is a right triangle.

NCERT Example 2 — show $(1,7),(4,2),(-1,-1),(-4,4)$ are a square

Let $A(1,7),B(4,2),C(-1,-1),D(-4,4)$. Sides:

$AB=\sqrt{(1-4)^{2}+(7-2)^{2}}=\sqrt{9+25}=\sqrt{34}$; $BC=\sqrt{25+9}=\sqrt{34}$;

$CD=\sqrt{9+25}=\sqrt{34}$; $DA=\sqrt{25+9}=\sqrt{34}$.

Diagonals: $AC=\sqrt{(1+1)^{2}+(7+1)^{2}}=\sqrt{4+64}=\sqrt{68}$; $BD=\sqrt{64+4}=\sqrt{68}$.

All four sides equal and both diagonals equal $\Rightarrow$ $ABCD$ is a square.

NCERT Example 3 — are $A(3,1),B(6,4),C(8,6)$ in a line?

$AB=\sqrt{(6-3)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt2$;

$BC=\sqrt{(8-6)^{2}+(6-4)^{2}}=\sqrt{4+4}=2\sqrt2$;

$AC=\sqrt{(8-3)^{2}+(6-1)^{2}}=\sqrt{25+25}=5\sqrt2$.

Since $AB+BC=3\sqrt2+2\sqrt2=5\sqrt2=AC$, the points are collinear — they are seated in a line.

NCERT Example 4 — relation so $(x,y)$ is equidistant from $(7,1)$ and $(3,5)$

Let $P(x,y)$ with $PA=PB$, so $PA^{2}=PB^{2}$:

$(x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}$.

Expand: $x^{2}-14x+49+y^{2}-2y+1=x^{2}-6x+9+y^{2}-10y+25$.

Simplify: $-14x-2y+50=-6x-10y+34\Rightarrow x-y=2$. This line is the perpendicular bisector of $AB$.

NCERT Example 5 — point on the $y$-axis equidistant from $(6,5)$ and $(-4,3)$

A point on the $y$-axis is $P(0,y)$. Set $PA^{2}=PB^{2}$:

$(6-0)^{2}+(5-y)^{2}=(-4-0)^{2}+(3-y)^{2}$.

$36+25-10y+y^{2}=16+9-6y+y^{2}\Rightarrow 61-10y=25-6y\Rightarrow 4y=36\Rightarrow y=9$.

The point is $(0,9)$. Check: $AP=\sqrt{36+16}=\sqrt{52}$ and $BP=\sqrt{16+36}=\sqrt{52}$. ✓

5. The section formula (internal division)

The problem: point $P$ lies on segment $AB$ and divides it so that $PA:PB=m_1:m_2$. Where is $P$? Drawing perpendiculars and using similar triangles ($\triangle PAQ\sim\triangle BPC$ by AA) gives the result.

If $P(x,y)$ divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m_1:m_2$, then:

$$P=\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\ \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$$

Memory aid (cross pattern): the first part of each numerator uses $m_1$ with the second point's coordinate, and the second part uses $m_2$ with the first point's coordinate. Keep the ratio in the right order: $m_1$ belongs to the part near $A$.

If the ratio is written as $k:1$, the formula simplifies neatly:

$$P=\left(\dfrac{kx_2+x_1}{k+1},\ \dfrac{ky_2+y_1}{k+1}\right)$$

This $k:1$ form is the quickest way to find an unknown ratio: set one coordinate equal to its known value and solve for $k$.

6. The midpoint formula

The midpoint divides a segment in the ratio $1:1$. Putting $m_1=m_2=1$ into the section formula:

$$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$$

In words: the midpoint's coordinates are simply the averages of the endpoints' coordinates. This is heavily used for parallelograms, because the diagonals of a parallelogram bisect each other — so the midpoint of one diagonal equals the midpoint of the other.

7. Worked Examples on the section formula

NCERT Example 6 — divide $(4,-3)$ and $(8,5)$ in ratio $3:1$

Here $m_1:m_2=3:1$. $x=\dfrac{3(8)+1(4)}{3+1}=\dfrac{28}{4}=7$, $y=\dfrac{3(5)+1(-3)}{3+1}=\dfrac{12}{4}=3$.

So the required point is $(7,3)$.

NCERT Example 7 — in what ratio does $(-4,6)$ divide $A(-6,10),B(3,-8)$?

Let the ratio be $k:1$. Using the $x$-coordinate: $-4=\dfrac{3k-6}{k+1}$.

$-4k-4=3k-6\Rightarrow 7k=2\Rightarrow k=\dfrac{2}{7}$, i.e. ratio $2:7$.

Verify with $y$: $\dfrac{-8(2)+10(7)}{2+7}=\dfrac{-16+70}{9}=\dfrac{54}{9}=6$. ✓ So $(-4,6)$ divides $AB$ in $2:7$.

NCERT Example 8 — points of trisection of $A(2,-2),B(-7,4)$

Trisection points $P,Q$ satisfy $AP=PQ=QB$. So $P$ divides $AB$ in $1:2$:

$P=\left(\dfrac{1(-7)+2(2)}{3},\dfrac{1(4)+2(-2)}{3}\right)=\left(\dfrac{-3}{3},\dfrac{0}{3}\right)=(-1,0)$.

$Q$ divides $AB$ in $2:1$:

$Q=\left(\dfrac{2(-7)+1(2)}{3},\dfrac{2(4)+1(-2)}{3}\right)=\left(\dfrac{-12}{3},\dfrac{6}{3}\right)=(-4,2)$.

The points of trisection are $(-1,0)$ and $(-4,2)$.

NCERT Example 9 — ratio in which the $y$-axis divides $(5,-6),(-1,-4)$

Let the ratio be $k:1$. The dividing point is $\left(\dfrac{-k+5}{k+1},\dfrac{-4k-6}{k+1}\right)$. On the $y$-axis the abscissa is $0$:

$\dfrac{-k+5}{k+1}=0\Rightarrow k=5$, so the ratio is $5:1$.

Point of intersection: $y=\dfrac{-4(5)-6}{5+1}=\dfrac{-26}{6}=\dfrac{-13}{3}$, giving $\left(0,\dfrac{-13}{3}\right)$.

NCERT Example 10 — find $p$ if $A(6,1),B(8,2),C(9,4),D(p,3)$ is a parallelogram

Diagonals of a parallelogram bisect each other, so midpoint of $AC$ = midpoint of $BD$:

$\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{2+3}{2}\right)\Rightarrow\left(\dfrac{15}{2},\dfrac{5}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{5}{2}\right)$.

So $\dfrac{15}{2}=\dfrac{8+p}{2}\Rightarrow 8+p=15\Rightarrow p=7$.

8. A note on external division

The section formula above is for internal division — $P$ lies between $A$ and $B$, with $PA:PB=m_1:m_2$. If $P$ lies on line $AB$ but outside the segment, that is external division (studied in higher classes). For Class 10 boards, always assume internal division unless told otherwise.

How to choose the right tool quickly. Read the wording of the question and match it to a formula. Phrases like "how far apart", "length of a side" or "find the distance" point straight to the distance formula. Phrases like "divides in the ratio", "point of trisection" or "divides into equal parts" call for the section formula. Words like "centre", "midpoint" or "diameter" mean the midpoint formula. When asked to "prove the shape is a square / rhombus / parallelogram", compute all four sides and both diagonals with the distance formula and compare. When asked to find an unknown ratio in which a point or an axis divides a segment, set the ratio as $k:1$, substitute into the section formula, and solve the single resulting equation. A quick rough sketch before you start almost always prevents sign errors and tells you which quadrant the final answer should lie in.

9. NCERT Exercise 7.1 — fully solved

Q1. Distance between pairs of points.

  • (i) $(2,3),(4,1)$: $\sqrt{(4-2)^{2}+(1-3)^{2}}=\sqrt{4+4}=2\sqrt2$.
  • (ii) $(-5,7),(-1,3)$: $\sqrt{(-1+5)^{2}+(3-7)^{2}}=\sqrt{16+16}=4\sqrt2$.
  • (iii) $(a,b),(-a,-b)$: $\sqrt{(-a-a)^{2}+(-b-b)^{2}}=\sqrt{4a^{2}+4b^{2}}=2\sqrt{a^{2}+b^{2}}$.

Q2. Distance between $(0,0)$ and $(36,15)$: $\sqrt{36^{2}+15^{2}}=\sqrt{1296+225}=\sqrt{1521}=39$. So towns $A$ and $B$ (from Section 7.2) are $39$ km apart.

Q3. Are $(1,5),(2,3),(-2,-11)$ collinear? $AB=\sqrt{1+4}=\sqrt5$; $BC=\sqrt{16+196}=\sqrt{212}$; $AC=\sqrt{9+256}=\sqrt{265}$. Since no two distances sum to the third, the points are not collinear.

Q4. Is $(5,-2),(6,4),(7,-2)$ isosceles? $AB=\sqrt{1+36}=\sqrt{37}$; $BC=\sqrt{1+36}=\sqrt{37}$; $AC=\sqrt{4+0}=2$. Since $AB=BC$, the triangle is isosceles.

Q5. Champa vs Chameli (Fig. 7.8): $A(3,4),B(6,7),C(9,4),D(6,1)$. $AB=\sqrt{9+9}=3\sqrt2$; $BC=\sqrt{9+9}=3\sqrt2$; $CD=\sqrt{9+9}=3\sqrt2$; $DA=\sqrt{9+9}=3\sqrt2$. Diagonals $AC=\sqrt{36+0}=6$ and $BD=\sqrt{0+36}=6$. All sides equal and both diagonals equal $\Rightarrow$ it is a square, so Champa is correct.

Q6. Name the quadrilateral.

  • (i) $(-1,-2),(1,0),(-1,2),(-3,0)$: each side $=2\sqrt2$; diagonals $AC=4$ and $BD=4$. All sides and diagonals equal $\Rightarrow$ square.
  • (ii) $(-3,5),(3,1),(0,3),(-1,-4)$: here three of the points turn out collinear, so no quadrilateral is formed.
  • (iii) $(4,5),(7,6),(4,3),(1,2)$: opposite sides $AB=CD=\sqrt{10}$ and $BC=DA=\sqrt{18}$; diagonals $AC=2$ and $BD=\sqrt{52}$ are unequal $\Rightarrow$ parallelogram.

Q7. Point on $x$-axis equidistant from $(2,-5)$ and $(-2,9)$. Take $(x,0)$: $(x-2)^{2}+25=(x+2)^{2}+81$. Expand: $-4x+29=4x+85\Rightarrow -8x=56\Rightarrow x=-7$. Point is $(-7,0)$.

Q8. Values of $y$ if distance $PQ=10$ for $P(2,-3),Q(10,y)$. $\sqrt{(10-2)^{2}+(y+3)^{2}}=10\Rightarrow 64+(y+3)^{2}=100\Rightarrow (y+3)^{2}=36\Rightarrow y+3=\pm6$. So $y=3$ or $y=-9$.

Q9. $Q(0,1)$ equidistant from $P(5,-3),R(x,6)$. $QP^{2}=QR^{2}$: $25+16=x^{2}+25\Rightarrow x^{2}=16\Rightarrow x=\pm4$. For $x=4$: $QR=\sqrt{16+25}=\sqrt{41}$, $PR=\sqrt{1+81}=\sqrt{82}$. For $x=-4$: $QR=\sqrt{16+25}=\sqrt{41}$, $PR=\sqrt{81+81}=9\sqrt2$.

Q10. Relation so $(x,y)$ is equidistant from $(3,6)$ and $(-3,4)$. $(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$. Expand and simplify: $-6x-12y+45=6x-8y+25\Rightarrow -12x-4y+20=0\Rightarrow 3x+y=5$.

10. NCERT Exercise 7.2 — fully solved

Q1. Point dividing $(-1,7),(4,-3)$ in ratio $2:3$: $x=\dfrac{2(4)+3(-1)}{5}=\dfrac{5}{5}=1$, $y=\dfrac{2(-3)+3(7)}{5}=\dfrac{15}{5}=3$. Point $(1,3)$.

Q2. Trisection points of $(4,-1),(-2,-3)$. $P$ in $1:2$: $\left(\dfrac{1(-2)+2(4)}{3},\dfrac{1(-3)+2(-1)}{3}\right)=\left(2,\dfrac{-5}{3}\right)$. $Q$ in $2:1$: $\left(\dfrac{2(-2)+1(4)}{3},\dfrac{2(-3)+1(-1)}{3}\right)=\left(0,\dfrac{-7}{3}\right)$.

Q3. Flags on the ground (Fig. 7.12). Niharika's green flag: $\tfrac14$ of $AD$ on the 2nd line $\Rightarrow (2,25)$. Preet's red flag: $\tfrac15$ of $AD$ on the 8th line $\Rightarrow (8,20)$. Distance between flags $=\sqrt{(8-2)^{2}+(20-25)^{2}}=\sqrt{36+25}=\sqrt{61}$ m. Blue flag at the midpoint: $\left(\dfrac{2+8}{2},\dfrac{25+20}{2}\right)=(5,22.5)$, i.e. on the 5th line at a distance of $22.5$ m.

Q4. Ratio in which $(-1,6)$ divides $(-3,10),(6,-8)$. Let ratio $k:1$ on the $x$-coordinate: $-1=\dfrac{6k-3}{k+1}\Rightarrow -k-1=6k-3\Rightarrow 7k=2\Rightarrow k=\dfrac27$. Ratio $2:7$.

Q5. Ratio in which the $x$-axis divides $A(1,-5),B(-4,5)$; find the point. Let ratio $k:1$; on the $x$-axis the ordinate is $0$: $\dfrac{5k-5}{k+1}=0\Rightarrow k=1$, ratio $1:1$. Point of division: $x=\dfrac{-4+1}{2}=\dfrac{-3}{2}$, $y=0$, i.e. $\left(\dfrac{-3}{2},0\right)$.

Q6. Parallelogram $(1,2),(4,y),(x,6),(3,5)$. Diagonals bisect: midpoint of $(1,2)$–$(x,6)$ = midpoint of $(4,y)$–$(3,5)$. So $\dfrac{1+x}{2}=\dfrac{4+3}{2}\Rightarrow x=6$, and $\dfrac{2+6}{2}=\dfrac{y+5}{2}\Rightarrow 8=y+5\Rightarrow y=3$. Thus $x=6,\ y=3$.

Q7. $AB$ is a diameter; centre $(2,-3)$, $B(1,4)$; find $A$. The centre is the midpoint of $AB$. Let $A(x,y)$: $\dfrac{x+1}{2}=2\Rightarrow x=3$; $\dfrac{y+4}{2}=-3\Rightarrow y=-10$. So $A(3,-10)$.

Q8. $A(-2,-2),B(2,-4)$, find $P$ on $AB$ with $AP=\dfrac37 AB$. Then $AP:PB=3:4$. $P=\left(\dfrac{3(2)+4(-2)}{7},\dfrac{3(-4)+4(-2)}{7}\right)=\left(\dfrac{-2}{7},\dfrac{-20}{7}\right)$.

Q9. Points dividing $A(-2,2),B(2,8)$ into four equal parts. The three points divide $AB$ in $1:3$, $1:1$ (midpoint) and $3:1$. $P_1=\left(\dfrac{1(2)+3(-2)}{4},\dfrac{1(8)+3(2)}{4}\right)=\left(-1,\dfrac72\right)$; midpoint $P_2=(0,5)$; $P_3=\left(\dfrac{3(2)+1(-2)}{4},\dfrac{3(8)+1(2)}{4}\right)=\left(1,\dfrac{13}{2}\right)$.

Q10. Area of the rhombus with vertices $(3,0),(4,5),(-1,4),(-2,-1)$. Area of a rhombus $=\dfrac12\times(\text{product of diagonals})$. Diagonals: $d_1=\sqrt{(3+1)^{2}+(0-4)^{2}}=\sqrt{16+16}=4\sqrt2$; $d_2=\sqrt{(4+2)^{2}+(5+1)^{2}}=\sqrt{36+36}=6\sqrt2$. Area $=\dfrac12(4\sqrt2)(6\sqrt2)=\dfrac12\times48=24$ square units.

11. Common mistakes to avoid

  • Forgetting to square both differences — writing $\sqrt{(x_2-x_1)+(y_2-y_1)}$ instead of $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$.
  • Mixing up the cross pattern in the section formula — pairing $m_1$ with $x_1$ instead of $x_2$.
  • Getting the ratio order wrong: $PA:PB=m_1:m_2$, with $m_1$ on the side of $A$.
  • For a square, checking only that all sides are equal — you must also check the diagonals are equal (otherwise it could be a rhombus).
  • Sign slips with negative coordinates: $(2-(-3))=5$, not $-1$.
  • For a point on the $x$-axis write $(x,0)$ and on the $y$-axis write $(0,y)$ — not the reverse.

12. Quick revision checklist

  • Distance $=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$; from origin $=\sqrt{x^{2}+y^{2}}$.
  • Section formula $\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$; use $k:1$ to find an unknown ratio.
  • Midpoint $=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ — averages of the coordinates.
  • Collinear $\Leftrightarrow$ largest distance $=$ sum of the other two.
  • Square: all sides equal + diagonals equal. Parallelogram: diagonals bisect (equal midpoints).
  • Equidistant condition: set $PA^{2}=PB^{2}$ and simplify to a linear relation.
Practice MCQs
1. The distance between the points $(2,3)$ and $(4,1)$ is:
  1. $2\sqrt2$
  2. $4$
  3. $2$
  4. $\sqrt8\,+\,1$
Answer: (A) $\sqrt{(4-2)^{2}+(1-3)^{2}}=\sqrt{4+4}=2\sqrt2$.
2. The distance of the point $(-6,8)$ from the origin is:
  1. $8$
  2. $2\sqrt7$
  3. $10$
  4. $14$
Answer: (C) $\sqrt{(-6)^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10$.
3. The midpoint of the segment joining $(2,-3)$ and $(6,7)$ is:
  1. $(4,2)$
  2. $(8,4)$
  3. $(4,4)$
  4. $(2,2)$
Answer: (A) $\left(\dfrac{2+6}{2},\dfrac{-3+7}{2}\right)=(4,2)$.
4. The point that divides $(4,-3)$ and $(8,5)$ in the ratio $3:1$ internally is:
  1. $(6,1)$
  2. $(7,3)$
  3. $(5,2)$
  4. $(7,4)$
Answer: (B) $x=\dfrac{3(8)+1(4)}{4}=7$, $y=\dfrac{3(5)+1(-3)}{4}=3$.
5. The points $A(3,1),B(6,4),C(8,6)$ are:
  1. vertices of a triangle
  2. collinear
  3. vertices of a square
  4. equidistant from origin
Answer: (B) $AB+BC=3\sqrt2+2\sqrt2=5\sqrt2=AC$, so collinear.
6. If the distance between $(2,-3)$ and $(10,y)$ is $10$ units, then $y=$
  1. $3$ only
  2. $-9$ only
  3. $3$ or $-9$
  4. $6$ or $-6$
Answer: (C) $64+(y+3)^{2}=100\Rightarrow (y+3)^{2}=36\Rightarrow y=3$ or $-9$.
7. The $y$-axis divides the segment joining $(5,-6)$ and $(-1,-4)$ in the ratio:
  1. $1:5$
  2. $5:1$
  3. $2:3$
  4. $3:2$
Answer: (B) Setting abscissa $0$ in $k:1$ form: $\dfrac{-k+5}{k+1}=0\Rightarrow k=5$, ratio $5:1$.
8. If $(1,2),(4,y),(x,6),(3,5)$ are vertices of a parallelogram (in order), then $(x,y)=$
  1. $(6,3)$
  2. $(3,6)$
  3. $(5,4)$
  4. $(4,5)$
Answer: (A) Diagonals bisect $\Rightarrow x=6,\ y=3$.
9. The points $(5,-2),(6,4),(7,-2)$ form a triangle that is:
  1. scalene
  2. isosceles
  3. equilateral
  4. right-angled
Answer: (B) $AB=BC=\sqrt{37}$, $AC=2$, so isosceles.
10. The coordinates of the point on the $x$-axis equidistant from $(2,-5)$ and $(-2,9)$ are:
  1. $(7,0)$
  2. $(-7,0)$
  3. $(0,7)$
  4. $(0,-7)$
Answer: (B) $(x-2)^{2}+25=(x+2)^{2}+81\Rightarrow x=-7$, point $(-7,0)$.
11. The distance between $(a,b)$ and $(-a,-b)$ is:
  1. $a+b$
  2. $2\sqrt{a^{2}+b^{2}}$
  3. $\sqrt{a^{2}+b^{2}}$
  4. $2(a+b)$
Answer: (B) $\sqrt{4a^{2}+4b^{2}}=2\sqrt{a^{2}+b^{2}}$.
12. The centre of a circle is the midpoint of a diameter $AB$. If the centre is $(2,-3)$ and $B(1,4)$, then $A=$
  1. $(3,-10)$
  2. $(1,-2)$
  3. $(3,1)$
  4. $(-1,-10)$
Answer: (A) $\dfrac{x+1}{2}=2,\ \dfrac{y+4}{2}=-3\Rightarrow A(3,-10)$.
Assertion–Reason
A: The midpoint of the segment joining $(1,7)$ and $(3,3)$ is $(2,5)$.   R: The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$.
Answer: Both A and R are true, and R is the correct explanation of A — $\left(\dfrac{1+3}{2},\dfrac{7+3}{2}\right)=(2,5)$.
A: The points $(1,5),(2,3)$ and $(-2,-11)$ are collinear.   R: Three points are collinear if the sum of two of the distances equals the third.
Answer: R is true, but A is false — here $AB=\sqrt5,\ BC=\sqrt{212},\ AC=\sqrt{265}$ and no two sum to the third, so the points are not collinear.
Previous-year questions
Q1. Find the ratio in which the point $(-4,6)$ divides the segment joining $A(-6,10)$ and $B(3,-8)$. (CBSE, 3 marks)
Answer: Take ratio $k:1$. From $x$: $-4=\dfrac{3k-6}{k+1}\Rightarrow 7k=2\Rightarrow k=\dfrac27$. Ratio $2:7$ (verified by the $y$-coordinate).
Q2. Show that the points $(1,7),(4,2),(-1,-1),(-4,4)$ are the vertices of a square. (CBSE, 3 marks)
Answer: All four sides $=\sqrt{34}$ and both diagonals $=\sqrt{68}$; equal sides and equal diagonals $\Rightarrow$ square.
Q3. Find the value of $y$ for which the distance between $P(2,-3)$ and $Q(10,y)$ is $10$ units. (CBSE, 2 marks)
Answer: $64+(y+3)^{2}=100\Rightarrow (y+3)^{2}=36\Rightarrow y=3$ or $y=-9$.
Q4. Find the coordinates of the points of trisection of the segment joining $A(2,-2)$ and $B(-7,4)$. (CBSE, 3 marks)
Answer: $P$ in $1:2$ gives $(-1,0)$; $Q$ in $2:1$ gives $(-4,2)$.
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More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Arithmetic Progressions · Triangles · Introduction to Trigonometry · Some Applications of Trigonometry