# CIE A level -Pure Mathematics 1 : Topic : 1.3 Coordinate geometry: interpret and use any of the forms y = mx + c : Exam Questions Paper 1

### Question

The coordinates of A are (-3,2) and the coordinates of C are (5,).The mid point of AC is M and the perpendicular bisector of AC cuts the x-axis at B.

(i)Find the equation of MB and the coordinates ofB.

(ii)Show that AB is perpendicular to BC.

(iii)Given that ABCD is a square, find the coordinates of D and the length of AD.

(i) Equation of $$MB$$ and Coordinates of $$B$$:
Given that $$A(-3,2)$$ and $$C(5,y_C)$$:
The midpoint $$M$$ is found as $$M\left(\frac{-3+5}{2}, \frac{2+y_C}{2}\right) = M(1, \frac{y_C + 2}{2})$$.
The slope of $$MB$$ is the negative reciprocal of the slope of $$AC$$, which is $$\frac{y_C – 2}{5 – (-3)} = \frac{y_C – 2}{8}$$.
Using the point-slope form of the equation, the equation of $$MB$$ is $$y – 4 = -\frac{y_C – 2}{8}(x – 1)$$.
For $$MB$$ to intersect the x-axis, $$y$$ must be $$0$$:
$$0 – 4 = -\frac{y_C – 2}{8}(x – 1)$$
Solving for $$x$$, we get $$x = 3$$. Therefore, the coordinates of $$B$$ are $$(3, 0)$$.
(ii) Proving $$AB$$ is Perpendicular to $$BC$$:
The slope of $$AB$$ is found using the coordinates of $$A$$ and $$B$$:
$$m_{AB} = \frac{0 – 2}{3 – (-3)} = -\frac{1}{2}$$
The slope of $$BC$$ is $$\frac{y_C – 2}{5 – (-3)} = \frac{y_C – 2}{8}$$.
To check if $$AB$$ is perpendicular to $$BC$$, we check if $$m_{AB} \cdot m_{BC} = -1$$:
$$-\frac{1}{2} \cdot \frac{y_C – 2}{8} = -1$$
$$y_C – 2 = 4$$
Solving for $$y_C$$, we get $$y_C = 6$$, confirming that $$AB$$ is perpendicular to $$BC$$.
(iii) Finding $$D$$ and Length of $$AD$$:
Since $$D$$ lies on the perpendicular bisector ($$x = 1$$), the coordinates of $$D$$ are $$(1, 2)$$. The length of $$AD$$ is then calculated using the distance formula:
$$AD = \sqrt{(1 – (-3))^2 + (2 – 2)^2} = \sqrt{40} \approx 6.32$$

### Question

C is the mid-point of the line joining A(14,-7) to B(-6,3).Tge line through C perpendicular to AB crosses the y-axis Aat D

(i)Find the equation of the line CD,giving your answer in the form y=mx+c

(i) Equation of Line CD:
Given that $$C$$ is the midpoint of $$AB$$, the coordinates of $$C$$ are $$\left(\frac{14 + (-6)}{2}, \frac{(-7) + 3}{2}\right) = (4, -2)$$.
The slope of $$AB$$ is $$\frac{3 – (-7)}{(-6) – 14} = \frac{10}{-20} = -\frac{1}{2}$$. Therefore, the slope of the line perpendicular to $$AB$$ (and hence the slope of $$CD$$) is $$2$$.
Using the point-slope form of the equation, the equation of $$CD$$ is $$y – (-2) = 2(x – 4)$$.
Simplifying, we get $$y + 2 = 2x – 8$$, and finally, $$y = 2x – 10$$.
The distance between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is given by the distance formula:
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
In this case, $$A(14, -7)$$ and $$D(x_D, 0)$$, where $$x_D$$ is the x-coordinate of $$D$$.
$$AD = \sqrt{(14-x_D)^2 + (-7-(-10))^2}$$
Therefore, the distance $$AD$$ is $$\sqrt{205}$$.

Question

The points A(1,1) and B(5,9) lie on the curve$$6y=5x^{2}-18x+19$$

(i)Show that the equation of the perpendicular bisector of AB is $$2y=13-x$$

The perpendicular bisector of AB meets the curve C and D.

(ii)Find ,by calculation ,the distance CD,giving your answer in the form $$\sqrt{\frac{p}{q}}$$,where p and q are integers.

(i)The midpoint of $$AB$$ is given by the average of the x-coordinates and y-coordinates of $$A$$ and $$B$$:
$$\left(\frac{1+5}{2}, \frac{1+9}{2}\right) = (3, 5)$$
The slope of $$AB$$ is given by:
$$m_{AB} = \frac{9 – 1}{5 – 1} = \frac{8}{4} = 2$$
The negative reciprocal of the slope of $$AB$$ (since the bisector is perpendicular) is $$m_{\text{bisector}} = -\frac{1}{2}$$.
Using the point-slope form, the equation of the perpendicular bisector is given by:
$$y – 5 = -\frac{1}{2}(x – 3)$$
Simplifying, we get $$2y – 10 = -x + 3$$, and finally, $$2y = 13 – x$$.
(ii)$$(x^2 – 3x – 4) = 0$$. Solving for $$x$$, you found $$x = 4$$ and $$x = -1$$.
Now, substitute these x-values back into the equation of the curve $$6y = 5x^2 – 18x + 19$$ to find the corresponding y-values:
For $$x = 4$$: $$6y = 5(4)^2 – 18(4) + 19$$, solving this gives $$y = 4 \frac{1}{2}$$.
For $$x = -1$$: $$6y = 5(-1)^2 – 18(-1) + 19$$, solving this gives $$y = 7$$.
So, $$C = (4, 5)$$ and $$D = (-1, 7)$$. Now, use the distance formula to find $$CD$$:
$$CD = \sqrt{(-1 – 4)^2 + (2\frac{1}{2})^2} = \sqrt{\frac{125}{4}}$$
Therefore, the distance $$CD$$ is $$\sqrt{\frac{125}{4}}$$.

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