Solved Questions

How to Use These Questions

These 31 questions come in mixed order, as they would in a paper: a reflection may follow a centroid, and an area may follow a pair of equations. In CGL 2025 Tier 1 most shifts had one question from this chapter, usually on slope, the midpoint, distance or the equation of a line. Before any working, sketch the axes and put the points roughly where they belong. The sketch shows which vertex is opposite which, whether a line rises or falls, and whether the sign of your answer is sensible.

Answers are left as exact numbers, fractions or surds. Only two are marked [Tier 2]: external division and the angle between two lines. The foot of the perpendicular and the distance questions near the end are Tier 1 work, so do not skip those. Most of the Tier 1 questions need four lines of working or fewer. If one took you much longer, find the matching method in Shortcuts & Speed Methods and try the question again a few days later.

Solved Examples

Question 1

What kind of triangle has the vertices $A(-3, 1)$, $B(1, 4)$ and $C(4, 0)$? Find its area too.

Answer. I find the three sides first. AB has differences 4 and 3, so $AB = 5$. BC has differences 3 and 4, so $BC = 5$. AC has differences 7 and 1, so $AC = \sqrt{50}$. Two sides are equal, and $5^2 + 5^2 = 50 = AC^2$, so the angle at B is a right angle. The triangle is right-angled and isosceles, with the right angle at B. Its area is half the product of the two legs: $\frac{1}{2} \times 5 \times 5 =$ 12.5 square units.

Question 2

The line through $(p, 7)$ and $(2, -2)$ has slope $-\frac{3}{2}$. Find p.

Answer. The slope is $\frac{-2 - 7}{2 - p} = \frac{-9}{2 - p}$. Setting it equal to $-\frac{3}{2}$ and cross-multiplying, $-18 = -3(2 - p)$, so $2 - p = 6$ and $p = -4$. To check: from $(-4, 7)$ to $(2, -2)$, y falls 9 while x rises 6, and $-\frac{9}{6} = -\frac{3}{2}$.

Question 3

The point $P(k, -3)$ lies on the segment joining $A(-5, -6)$ and $B(7, 3)$. Find the ratio in which P divides AB, and the value of k.

Answer. The y-coordinate of P is the one I know, so I use the y's: $AP : PB = (-3 + 6) : (3 + 3) = 3 : 6 =$ 1 : 2. So P is one-third of the way from A to B. The x-coordinate goes from $-5$ to 7, a rise of 12, and a third of 12 is 4, so $k = -5 + 4 =$ $-1$.

Question 4

Find the point on the x-axis that is equidistant from $A(-1, 5)$ and $B(7, 3)$.

Answer. A point on the x-axis is $P(x, 0)$. Setting $PA^2 = PB^2$: $(x + 1)^2 + 25 = (x - 7)^2 + 9$. The $x^2$ terms cancel and leave $2x + 26 = -14x + 58$, so $16x = 32$ and $x = 2$. The point is $(2, 0)$, and it is $\sqrt{9 + 25} = \sqrt{34}$ units from each of A and B.

Question 5

Find the equation of the line through $(-2, 7)$ and $(4, -5)$, and the point where it cuts the x-axis.

Answer. The slope is $\frac{-5 - 7}{4 + 2} = -2$. Point-slope form at $(-2, 7)$: $y - 7 = -2(x + 2)$, so $y = -2x + 3$, that is $2x + y = 3$. The other point checks: $8 - 5 = 3$. Putting $y = 0$ gives $x = \frac{3}{2}$, so the line cuts the x-axis at $(\frac{3}{2}, 0)$.

Question 6

The point $A(-3, 5)$ is reflected in the y-axis to B, and B is reflected in the origin to C. Find the area of triangle ABC.

Answer. In the y-axis only the sign of x changes, so $B(3, 5)$. In the origin both signs change, so $C(-3, -5)$. A and B have the same y, so AB is level, and A and C have the same x, so AC is upright. The right angle is at A, with $AB = 6$ and $AC = 10$. The area is $\frac{1}{2} \times 6 \times 10 =$ 30 square units.

Question 7

The vertices of a triangle are $A(1, 6)$, $B(-5, 2)$ and $C(7, -2)$. Find the length of the median through A, and the distance of the centroid from A.

Answer. The median from A goes to the midpoint of BC, which is $M(\frac{-5 + 7}{2}, \frac{2 - 2}{2}) = (1, 0)$. A and M have the same x, so $AM = 6 - 0 =$ 6 units. The centroid is $G(\frac{3}{3}, \frac{6}{3}) = (1, 2)$, so $AG = 6 - 2 =$ 4 units. That is two-thirds of the median, which is what it should be.

Question 8

For what value of k is the line $3x + ky = 5$ perpendicular to the line $4x - 6y + 1 = 0$?

Answer. Lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are perpendicular when $a_1a_2 + b_1b_2 = 0$; this is $m_1m_2 = -1$ with the fractions cleared. Here $3 \times 4 + k \times (-6) = 0$, so $6k = 12$ and $k = 2$. To check, the slopes are $-\frac{3}{2}$ and $\frac{2}{3}$, and their product is $-1$.

Question 9

Find the value of k for which $2x + ky = 6$ and $kx + 8y = 12$ are the same line. What happens for the other value of k that makes $\frac{a_1}{a_2} = \frac{b_1}{b_2}$?

Answer. $\frac{2}{k} = \frac{k}{8}$ gives $k^2 = 16$, so k is 4 or $-4$. For the same line the constants must give the same ratio too, and $\frac{6}{12} = \frac{1}{2}$. With $k = 4$ the ratios are $\frac{2}{4}$, $\frac{4}{8}$ and $\frac{6}{12}$, all equal to $\frac{1}{2}$, so the lines coincide: $k = 4$. With $k = -4$ the first two ratios are $-\frac{1}{2}$ but the third is $\frac{1}{2}$, so the lines are parallel, and the pair has no solution.

Question 10

Find the values of a for which the distance between $(a, -1)$ and $(3, 5)$ is 10 units.

Answer. The y-difference is 6, so the x-difference must complete a 6-8-10 triangle: $|a - 3| = 8$. So $a - 3 = 8$ or $a - 3 = -8$, which gives $a = 11$ or $a = -5$. Both points are 10 units from $(3, 5)$, one on each side of it.

Question 11

The graph shows a line L. Find its equation, and say whether the point $(6, 6)$ lies on it.

Answer. L cuts the x-axis at $(-3, 0)$ and the y-axis at $(0, 2)$, so in intercept form it is $\frac{x}{-3} + \frac{y}{2} = 1$. Multiplying by 6 gives $-2x + 3y = 6$, so the line is $2x - 3y + 6 = 0$. For $(6, 6)$: $12 - 18 + 6 = 0$, so yes, it lies on L. The graph agrees: the line goes up 2 for every 3 across, a slope of $\frac{2}{3}$, and $-\frac{a}{b} = -\frac{2}{-3} = \frac{2}{3}$.

Question 12

$A(-3, -2)$, $B(2, -4)$ and $C(5, 1)$ are three vertices of parallelogram ABCD, taken in order. Find D and the area of the parallelogram.

Answer. $D = A + C - B = (-3 + 5 - 2, -2 + 1 + 4) =$ $(0, 3)$. The diagonal AC cuts the parallelogram into two equal triangles, so its area is twice the area of triangle ABC. I move A to the origin: B becomes $(5, -2)$ and C becomes $(8, 3)$. The area of ABC is $\frac{1}{2}|5 \times 3 - 8 \times (-2)| = \frac{31}{2}$, so the parallelogram is 31 square units.

Question 13

Find the image of $P(5, -1)$ in the line $x = -2$, and then the image of that point in the line $y = 3$.

Answer. The line $x = -2$ is upright, so y stays $-1$ and x becomes $2 \times (-2) - 5 = -9$. The first image is $(-9, -1)$; P was 7 units to the right of the line and this point is 7 units to its left. The line $y = 3$ is level, so x stays $-9$ and y becomes $2 \times 3 - (-1) = 7$. The final point is $(-9, 7)$.

Question 14

Find the equation of the perpendicular bisector of the segment joining $A(-5, 7)$ and $B(3, 3)$.

Answer. It passes through the midpoint of AB, $(-1, 5)$. The slope of AB is $\frac{3 - 7}{3 + 5} = -\frac{1}{2}$, so the bisector has slope 2. Then $y - 5 = 2(x + 1)$, that is $y = 2x + 7$, or $2x - y + 7 = 0$. To check, take another point on it, $(0, 7)$: it is $\sqrt{25 + 0} = 5$ from A and $\sqrt{9 + 16} = 5$ from B.

Question 15

Find the area of the region bounded by the coordinate axes, the line $5x + 2y = 20$ and the line $x = 2$.

Answer. I find the corners. The axes meet at $(0, 0)$, and $x = 2$ meets the x-axis at $(2, 0)$. It meets $5x + 2y = 20$ where $2y = 10$, at $(2, 5)$, and $5x + 2y = 20$ meets the y-axis at $(0, 10)$. The region is a trapezium with upright parallel sides 10 and 5, a distance 2 apart. Area $= \frac{1}{2}(10 + 5) \times 2 =$ 15 square units.

Question 16

Find the points that divide the segment joining $A(2, -4)$ and $B(-6, 8)$ into four equal parts.

Answer. From A to B, x changes by $-8$ and y by 12. A quarter of that is a step of $-2$ in x and 3 in y. Walking from A one step at a time gives $(0, -1)$, $(-2, 2)$ and $(-4, 5)$. The middle one is the midpoint of AB, and a fourth step lands on $(-6, 8)$, which is B, so the steps are right.

Question 17

A line passes through $(6, -2)$, and its x-intercept is twice its y-intercept. Neither intercept is zero. Find the equation of the line.

Answer. Let the y-intercept be b. Then the x-intercept is 2b, and the line is $\frac{x}{2b} + \frac{y}{b} = 1$, or $x + 2y = 2b$. Putting in $(6, -2)$: $6 - 4 = 2b$, so $b = 1$. The line is $x + 2y = 2$, with x-intercept 2 and y-intercept 1.

Question 18

Without solving, say whether each pair has one solution, no solution or infinitely many. Where there is exactly one, find it.

• (i) $6x - 4y = 14$ and $9x - 6y = 21$
• (ii) $7x + 3y = 4$ and $14x + 6y = 9$
• (iii) $x - 3y = 2$ and $3x - y = 6$

Answer. (i) The ratios are $\frac{6}{9}$, $\frac{-4}{-6}$ and $\frac{14}{21}$, all equal to $\frac{2}{3}$: the same line, so infinitely many solutions. (ii) $\frac{7}{14} = \frac{3}{6}$, but $\frac{4}{9}$ is different: parallel lines, so no solution. (iii) $\frac{1}{3}$ and $\frac{-3}{-1} = 3$ differ, so the lines cross at one point. From the first equation $x = 3y + 2$; then $9y + 6 - y = 6$ gives $y = 0$ and $x = 2$. They meet at $(2, 0)$.

Question 19

Show that $A(-2, -1)$, $B(2, 1)$, $C(0, 5)$ and $D(-4, 3)$ are the vertices of a square, and find its area.

Answer. Sides first. AB has differences 4 and 2, BC has 2 and 4, CD has 4 and 2, and DA has 2 and 4, so all four sides are $\sqrt{20}$. Then the diagonals: AC has differences 2 and 6, and BD has 6 and 2, so both are $\sqrt{40}$. Four equal sides make ABCD a rhombus, and a rhombus with equal diagonals is a square. Its area is $(\sqrt{20})^2 =$ 20 square units.

Question 20

The area of the triangle with vertices $(k, 0)$, $(4, 0)$ and $(0, 2)$ is 6 square units. Find k.

Answer. Two vertices are on the x-axis, so the base is $|4 - k|$, and the height is 2, the distance of $(0, 2)$ from the x-axis. Then $\frac{1}{2} \times |4 - k| \times 2 = 6$, so $|4 - k| = 6$. That gives $k = -2$ or $k = 10$: the third vertex can be on either side of $(4, 0)$.

Question 21

$A(-4, -1)$, $B(2, 2)$ and $C(12, k)$ are collinear. Find k, and the ratio in which B divides AC.

Answer. From A to B, x rises 6 and y rises 3, so y rises half as much as x. From B to C, x rises 10, so y rises 5, and $k = 7$. For the ratio I use the x's: $AB : BC = (2 + 4) : (12 - 2) = 6 : 10 =$ 3 : 5.

Question 22

The lines $3x - y = 3$ and $x + 3y = 21$ meet at a point P. Find P, and the area of the triangle the two lines make with the y-axis.

Answer. From the first equation $y = 3x - 3$. Putting this into the second: $x + 9x - 9 = 21$, so $x = 3$ and $y = 6$, and the point is $P(3, 6)$. The y-intercepts are $-3$ and 7, so the base on the y-axis is 10. The height is the distance of P from the y-axis, 3. Area $= \frac{1}{2} \times 10 \times 3 =$ 15 square units.

Question 23

Find the area of the quadrilateral with vertices $A(-4, -2)$, $B(3, -3)$, $C(5, 4)$ and $D(-1, 5)$, taken in order.

Answer. I cut it along AC into triangles ABC and ACD, and move A to the origin, so B, C and D become $(7, -1)$, $(9, 6)$ and $(3, 7)$.

• Triangle ABC: $\frac{1}{2}|7 \times 6 - 9 \times (-1)| = \frac{51}{2} = 25.5$
• Triangle ACD: $\frac{1}{2}|9 \times 7 - 3 \times 6| = \frac{45}{2} = 22.5$

The quadrilateral is $25.5 + 22.5 =$ 48 square units.

Question 24

Find the equation of the line through the point where $x + 2y = 9$ and $4x - y = 18$ meet, perpendicular to $3x - y = 4$.

Answer. From the second, $y = 4x - 18$; then $x + 8x - 36 = 9$, so $x = 5$ and $y = 2$. A line perpendicular to $3x - y = 4$ has the form $x + 3y = k$: swap the coefficients and change one sign. At $(5, 2)$, $k = 5 + 6 = 11$. The line is $x + 3y = 11$.

Question 25

The midpoints of the sides BC, CA and AB of triangle ABC are $D(3, 1)$, $E(5, 7)$ and $F(-3, 2)$. Find the vertices of the triangle.

Answer. Each side of triangle DEF is parallel to a side of ABC, so AFDE is a parallelogram with A opposite D. By the fourth-vertex rule, $A = E + F - D = (5 - 3 - 3, 7 + 2 - 1) = (-1, 8)$. In the same way $B = D + F - E = (-5, -4)$ and $C = D + E - F = (11, 6)$. The vertices are $A(-1, 8)$, $B(-5, -4)$ and $C(11, 6)$. One check: the midpoint of BC is $(\frac{6}{2}, \frac{2}{2}) = (3, 1)$, which is D.

Question 26

Find the point that is equidistant from $(2, 6)$, $(-5, 5)$ and $(3, -1)$, and the common distance.

Answer. Call the point $(x, y)$. Equal distances from $(2, 6)$ and $(-5, 5)$: after squaring, the $x^2$ and $y^2$ cancel and leave $-4x - 12y + 40 = 10x - 10y + 50$, which is $7x + y = -5$. Equal distances from $(2, 6)$ and $(3, -1)$: $-4x - 12y + 40 = -6x + 2y + 10$, which is $x - 7y = -15$. From the second, $x = 7y - 15$; then $49y - 105 + y = -5$, so $y = 2$ and $x = -1$. The point is $(-1, 2)$. Its differences from $(2, 6)$ are 3 and 4, so the common distance is 5 units.

Question 27

Find the area of the triangle formed by the lines $y = 1$, $3x + 4y = 34$ and $3x - 2y + 8 = 0$.

Answer. I find the three corners. $y = 1$ with $3x + 4y = 34$ gives $3x = 30$, so $(10, 1)$. $y = 1$ with $3x - 2y + 8 = 0$ gives $3x = -6$, so $(-2, 1)$. For the other two lines, taking $3x - 2y = -8$ from $3x + 4y = 34$ leaves $6y = 42$, so $y = 7$, $x = 2$, and the corner is $(2, 7)$. The base along $y = 1$ runs from $-2$ to 10, so it is 12, and $(2, 7)$ is $7 - 1 = 6$ above it. Area $= \frac{1}{2} \times 12 \times 6 =$ 36 square units.

Question 28

Find the foot of the perpendicular from $P(9, 2)$ to the line $4x - 3y = 5$. Find also the length of this perpendicular and the image of P in the line.

Answer. The perpendicular through P has the form $3x + 4y = k$, and P gives $k = 27 + 8 = 35$. Now I solve $4x - 3y = 5$ with $3x + 4y = 35$: multiplying the first by 4 and the second by 3 and adding gives $25x = 125$, so $x = 5$ and then $y = 5$. The foot is $F(5, 5)$. PF has differences 4 and 3, so its length is 5 units. The image is as far beyond F as P is before it, so $P' = 2F - P = (10 - 9, 10 - 2) =$ $(1, 8)$.

Question 29

[Tier 2] Find the point that divides the segment joining $A(3, -2)$ and $B(5, 4)$ externally in the ratio 2 : 1.

Answer. For external division in the ratio $m : n$ the section formula takes a minus sign: $(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n})$. With $m = 2$ and $n = 1$ this is $(\frac{10 - 3}{1}, \frac{8 + 2}{1}) =$ $(7, 10)$. Seen another way, the point is twice as far from A as from B, on the far side of B, so B is the midpoint of A and the point. Indeed $\frac{3 + 7}{2} = 5$ and $\frac{-2 + 10}{2} = 4$.

Question 30

Find the distance of the point $(3, -2)$ from the line $5x - 12y + 6 = 0$. Find also the distance between the parallel lines $3x + 4y = 7$ and $6x + 8y + 11 = 0$.

Answer. The distance of $(x_1, y_1)$ from $ax + by + c = 0$ is $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. Here that is $\frac{|15 + 24 + 6|}{\sqrt{25 + 144}} =$ $\frac{45}{13}$ units. For the parallel lines, first make the x and y coefficients match: halving the second gives $3x + 4y + 5.5 = 0$, and the first is $3x + 4y - 7 = 0$. The distance is the gap between the constants over $\sqrt{a^2 + b^2}$: $\frac{|-7 - 5.5|}{\sqrt{9 + 16}} = \frac{12.5}{5} =$ 2.5 units.

Question 31

[Tier 2] Find the acute angle between the lines $y = 3x + 1$ and $x - 2y + 9 = 0$.

Answer. The slopes are 3 and $\frac{1}{2}$. For the acute angle $\theta$ between two lines, $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$. Here $\tan\theta = \frac{3 - \frac{1}{2}}{1 + \frac{3}{2}} = \frac{\frac{5}{2}}{\frac{5}{2}} = 1$, so $\theta = 45^\circ$.