Class 10 Coordinate Geometry Practice Questions with Solutions

Very short answer (1 mark)#

Ques­tion 1#

Find the dis­tance between $(-6, -1)$ and $(6, 4)$.

$$\sqrt{(6 + 6)^2 + (4 + 1)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$

Answer: 13 units.

Ques­tion 2#

How far is the point $(7, -24)$ from the ori­gin?

$$\sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$

Answer: 25 units.

Ques­tion 3#

Find the mid­point of the seg­ment join­ing $(-5, 8)$ and $(9, -2)$.

$$\displaystyle \left( \frac{-5 + 9}{2},\ \frac{8 + (-2)}{2} \right) = (2, 3)$$

Answer: $(2, 3)$.

Ques­tion 4#

The point $\displaystyle \left( \frac{a}{2}, 3 \right)$ is the mid­point of the seg­ment join­ing $(-7, 1)$ and $(3, 5)$. Find $a$.

The mid­point is $\displaystyle \left( \frac{-7 + 3}{2}, \frac{1 + 5}{2} \right) = (-2, 3)$. The y-coor­di­nates agree. Equat­ing the x-coor­di­nates:

$$\displaystyle \frac{a}{2} = -2 \Rightarrow a = -4$$

Answer: $a = -4$.

Short answer (2 marks)#

Ques­tion 5#

Find the ratio in which the x-axis divides the seg­ment join­ing $(-3, -6)$ and $(5, 3)$. Also find the point of divi­sion.

Let the ratio be $k : 1$. On the x-axis, $y = 0$:

$$\displaystyle \frac{3k + (-6)}{k + 1} = 0 \Rightarrow 3k = 6 \Rightarrow k = 2$$

So the ratio is $2 : 1$. Then

$$\displaystyle x = \frac{2 \times 5 + 1 \times (-3)}{3} = \frac{7}{3}$$

Answer: $2 : 1$, at the point $\displaystyle \left( \frac{7}{3}, 0 \right)$.

Ques­tion 6#

Show that $A(0, -2)$, $B(3, 2)$ and $C(9, 10)$ are collinear.

$$\begin{aligned} AB &= \sqrt{3^2 + 4^2} = 5 \\ BC &= \sqrt{6^2 + 8^2} = 10 \\ AC &= \sqrt{9^2 + 12^2} = \sqrt{225} = 15 \end{aligned}$$

Since $AB + BC = 15 = AC$, the three points lie on one line.

Answer: The points are collinear, with $B$ between $A$ and $C$.

Why this works: if $B$ were off the line, $A$, $B$, $C$ would form a tri­an­gle and $AB + BC$ would be greater than $AC$.

Ques­tion 7#

Find $x$ if $P(x, 2)$ is equidis­tant from $A(3, -4)$ and $B(-1, 6)$.

$$\displaystyle \begin{aligned} PA^2 &= PB^2 \\ (x - 3)^2 + (2 + 4)^2 &= (x + 1)^2 + (2 - 6)^2 \\ x^2 - 6x + 9 + 36 &= x^2 + 2x + 1 + 16 \\ -6x + 45 &= 2x + 17 \\ 8x &= 28 \\ x &= \frac{7}{2} \end{aligned}$$

Answer: $\displaystyle x = \frac{7}{2}$.

Ques­tion 8#

Find the rela­tion between $x$ and $y$ if the point $(x, y)$ is equidis­tant from $(5, 1)$ and $(-1, 5)$.

$$\begin{aligned} (x - 5)^2 + (y - 1)^2 &= (x + 1)^2 + (y - 5)^2 \\ x^2 - 10x + 25 + y^2 - 2y + 1 &= x^2 + 2x + 1 + y^2 - 10y + 25 \\ -10x - 2y &= 2x - 10y \\ 8y &= 12x \\ 2y &= 3x \end{aligned}$$

Answer: $3x = 2y$, that is $3x - 2y = 0$.

Short answer (3 marks)#

Ques­tion 9#

Find the ratio in which $P(3, m)$ divides the seg­ment join­ing $A(-1, 7)$ and $B(5, -2)$. Hence find $m$.

Let the ratio be $k : 1$. Using the x-coor­di­nate:

$$\displaystyle \begin{aligned} \frac{5k + (-1)}{k + 1} &= 3 \\ 5k - 1 &= 3k + 3 \\ 2k &= 4 \\ k &= 2 \end{aligned}$$

The ratio is $2 : 1$. Using it in the y-coor­di­nate:

$$\displaystyle m = \frac{2 \times (-2) + 1 \times 7}{2 + 1} = \frac{3}{3} = 1$$

Answer: Ratio $2 : 1$; $m = 1$.

Ques­tion 10#

Find the points of tri­sec­tion of the seg­ment join­ing $(-4, 7)$ and $(5, -2)$.

The first point divides the seg­ment in $1 : 2$:

$$\displaystyle \left( \frac{1 \times 5 + 2 \times (-4)}{3},\ \frac{1 \times (-2) + 2 \times 7}{3} \right) = \left( \frac{-3}{3}, \frac{12}{3} \right) = (-1, 4)$$

The sec­ond divides it in $2 : 1$:

$$\displaystyle \left( \frac{2 \times 5 + 1 \times (-4)}{3},\ \frac{2 \times (-2) + 1 \times 7}{3} \right) = \left( \frac{6}{3}, \frac{3}{3} \right) = (2, 1)$$

Answer: $(-1, 4)$ and $(2, 1)$.

Why this works: each step along the seg­ment adds one third of the change, $\displaystyle \frac{1}{3}(9, -9) = (3, -3)$.

Long answer (4 to 5 marks)#

Ques­tion 11#

Show that $A(-2, -1)$, $B(1, 0)$, $C(4, 3)$ and $D(1, 2)$, taken in order, form a par­al­lel­o­gram but not a rec­tan­gle.

Sides:

$$\begin{aligned} AB &= \sqrt{3^2 + 1^2} = \sqrt{10} \\ BC &= \sqrt{3^2 + 3^2} = \sqrt{18} \\ CD &= \sqrt{(-3)^2 + (-1)^2} = \sqrt{10} \\ DA &= \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} \end{aligned}$$

Oppo­site sides are equal ($AB = CD$ and $BC = DA$), so $ABCD$ is a par­al­lel­o­gram.

Diag­o­nals:

$$AC = \sqrt{6^2 + 4^2} = \sqrt{52}, \qquad BD = \sqrt{0^2 + 2^2} = 2$$

The diag­o­nals are unequal, so the par­al­lel­o­gram is not a rec­tan­gle. As a fur­ther check, both diag­o­nals have mid­point $\left( 1, 1 \right)$.

Answer: $ABCD$ is a par­al­lel­o­gram with unequal diag­o­nals, so it is not a rec­tan­gle.

Ques­tion 12#

Show that $A(2, -1)$, $B(5, 3)$, $C(2, 7)$ and $D(-1, 3)$, taken in order, form a rhom­bus. Is it a square? Find its area.

$$\begin{aligned} AB &= \sqrt{3^2 + 4^2} = 5 \\ BC &= \sqrt{(-3)^2 + 4^2} = 5 \\ CD &= \sqrt{(-3)^2 + (-4)^2} = 5 \\ DA &= \sqrt{3^2 + (-4)^2} = 5 \end{aligned}$$

All four sides are equal, so $ABCD$ is a rhom­bus.

$$AC = \sqrt{0^2 + 8^2} = 8, \qquad BD = \sqrt{(-6)^2 + 0^2} = 6$$

The diag­o­nals are unequal, so it is not a square. The area of a rhom­bus is half the prod­uct of its diag­o­nals:

$$\displaystyle \text{Area} = \frac{1}{2} \times 8 \times 6 = 24$$

Answer: A rhom­bus, not a square; area 24 square units.

Ques­tion 13#

Show that $A(-1, 2)$, $B(3, -1)$ and $C(5, 10)$ are the ver­tices of a right-angled tri­an­gle. Name the right angle and state the type of tri­an­gle by sides.

$$\begin{aligned} AB &= \sqrt{4^2 + (-3)^2} = 5 \\ AC &= \sqrt{6^2 + 8^2} = 10 \\ BC &= \sqrt{2^2 + 11^2} = \sqrt{125} = 5\sqrt{5} \end{aligned}$$

Now $AB^2 + AC^2 = 25 + 100 = 125 = BC^2$. By the con­verse of Pythago­ras, the angle oppo­site $BC$, which is $\angle A$, is $90^\circ$. No two sides are equal.

Answer: A sca­lene right-angled tri­an­gle with $\angle A = 90^\circ$.

Ques­tion 14#

$A(2, 1)$, $B(5, y)$, $C(x, 7)$ and $D(1, 4)$ are the ver­tices of a par­al­lel­o­gram $ABCD$, taken in order. Find $x$ and $y$.

The diag­o­nals $AC$ and $BD$ bisect each other, so their mid­points coin­cide:

$$\displaystyle \left( \frac{2 + x}{2},\ \frac{1 + 7}{2} \right) = \left( \frac{5 + 1}{2},\ \frac{y + 4}{2} \right)$$

$$\begin{aligned} 2 + x &= 6 \Rightarrow x = 4 \\ 8 &= y + 4 \Rightarrow y = 4 \end{aligned}$$

Answer: $x = 4$, $y = 4$.

Ques­tion 15#

A line cuts the x-axis at $A$ and the y-axis at $B$. The mid­point of $AB$ is $M(2, -3)$. Find $A$, $B$ and the length $AB$.

Let $A = (a, 0)$ and $B = (0, b)$. Then

$$\displaystyle \left( \frac{a + 0}{2},\ \frac{0 + b}{2} \right) = (2, -3) \Rightarrow a = 4,\ b = -6$$

So $A = (4, 0)$ and $B = (0, -6)$.

$$AB = \sqrt{(0 - 4)^2 + (-6 - 0)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$

Answer: $A(4, 0)$, $B(0, -6)$ and $AB = 2\sqrt{13}$ units.

Ques­tion 16#

Find the points that divide the seg­ment join­ing $A(-3, 1)$ and $B(5, -7)$ into four equal parts.

The three points divide $AB$ in the ratios $1 : 3$, $1 : 1$ and $3 : 1$.

$$\displaystyle \begin{aligned} P_1 &= \left( \frac{1 \times 5 + 3 \times (-3)}{4},\ \frac{1 \times (-7) + 3 \times 1}{4} \right) = (-1, -1) \\ P_2 &= \left( \frac{-3 + 5}{2},\ \frac{1 + (-7)}{2} \right) = (1, -3) \\ P_3 &= \left( \frac{3 \times 5 + 1 \times (-3)}{4},\ \frac{3 \times (-7) + 1 \times 1}{4} \right) = (3, -5) \end{aligned}$$

Answer: $(-1, -1)$, $(1, -3)$ and $(3, -5)$.

Why this works: the mid­dle point is sim­ply the mid­point of $AB$, and the other two are the mid­points of $AP_2$ and $P_2B$.

Ques­tion 17#

Find the cen­troid of the tri­an­gle with ver­tices $(3, -5)$, $(-7, 4)$ and $(10, -2)$.

$$\displaystyle G = \left( \frac{3 + (-7) + 10}{3},\ \frac{-5 + 4 + (-2)}{3} \right) = \left( \frac{6}{3}, \frac{-3}{3} \right) = (2, -1)$$

Answer: $(2, -1)$.

Why this works: the cen­troid divides each median in the ratio $2 : 1$ from the ver­tex, and apply­ing the sec­tion for­mula to any median gives the aver­age of the three ver­tices.

Key terms#

Dis­tance for­mula

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, the length of the seg­ment join­ing two points.

Collinear

Lying on a sin­gle straight line.

Sec­tion for­mula

Gives the point divid­ing a seg­ment inter­nally in a given ratio.

Mid­point

The point that divides a seg­ment in the ratio $1 : 1$.

Points of tri­sec­tion

The two points that split a seg­ment into three equal parts.

Cen­troid

The point where the medi­ans of a tri­an­gle meet.

Rhom­bus

A quadri­lat­eral with four equal sides; its area is half the prod­uct of its diag­o­nals.

Con­verse of Pythago­ras

If the square of one side equals the sum of the squares of the other two, the angle oppo­site that side is a right angle.

Com­mon ques­tions#

How much work­ing should I show in a shape ques­tion?#

Show all four sides and both diag­o­nals, then state the prop­erty you are using, such as "oppo­site sides are equal and the diag­o­nals are unequal". The con­clu­sion earns marks only with its rea­son.

Which is quicker for collinear­ity, dis­tances or the sec­tion for­mula?#

Dis­tances always work. If one point looks like a mid­point or a tri­sec­tion point, check­ing it with the sec­tion for­mula can be quicker.

Why do we use the diag­o­nals, not the sides, to find a miss­ing ver­tex?#

The diag­o­nals of a par­al­lel­o­gram bisect each other, which gives two sim­ple lin­ear equa­tions. Equat­ing side lengths would give squared equa­tions that are harder to solve.

Should I sim­plify surds in the final answer?#

Yes. Write $\sqrt{52}$ as $2\sqrt{13}$ and $\sqrt{125}$ as $5\sqrt{5}$, unless the answer is only needed for a com­par­i­son.

Can I leave a ratio as a frac­tion?#

Write the final ratio in whole num­bers. A value $\displaystyle k = \frac{2}{3}$ means the ratio $2 : 3$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.