# CIE A level -Pure Mathematics 1 : Topic : 1.3 Coordinate geometry: equation of a straight line : 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) M=(1,4)  $$gradient=\frac{1}{2}$$

Equation of MB:$$y-4=-2\left ( x-1 \right )$$

When y=0,x=3 or B=(3,0)

(ii)Gradient of $$AB=-\frac{2}{6}$$;Gradient  of $$BC=\frac{6}{2}$$

$$m_{1}m_{2}=-1\left ( \Rightarrow AB perpendicular AC\right )$$

(iii) D=(-1,8) $$AD=\sqrt{40}$$ OR 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 the Line $$CD$$:
The midpoint $$C$$ of the line segment $$\overline{AB}$$ is given by the average of the corresponding coordinates of $$A$$ and $$B$$:
$$C\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$$
For points $$A(14, -7)$$ and $$B(-6, 3)$$, the coordinates of $$C$$ are:
$$C\left(\frac{14 + (-6)}{2}, \frac{(-7) + 3}{2}\right) = (4, -2)$$
Now, we know that the line $$CD$$ is perpendicular to $$\overline{AB}$$, so the slope of $$CD$$ is the negative reciprocal of the slope of $$\overline{AB}$$.
The slope of $$\overline{AB}$$ is given by:
$$m_{AB} = \frac{y_B – y_A}{x_B – x_A}$$
$$m_{AB} = \frac{3 – (-7)}{(-6) – 14} = \frac{10}{-20} = -\frac{1}{2}$$
The slope of $$CD$$ is the negative reciprocal:
$$m_{CD} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{2}} = 2$$
Now, by point-slope form of the equation to find the equation of $$CD$$:
$$y – y_C = m_{CD}(x – x_C)$$
$$y – (-2) = 2(x – 4)$$
$$y + 2 = 2x – 8$$
$$y = 2x – 10$$
So, the equation of the line $$CD$$ is $$y = 2x – 10$$.
(ii) The distance $$AD$$ is the length of the line segment $$\overline{AD}$$ and can be found using the distance formula:
$$AD = \sqrt{(x_D – x_A)^2 + (y_D – y_A)^2}$$
For $$A(14, -7)$$ and $$D(x_D, 0)$$ (since it lies on the y-axis), the distance $$AD$$ is:
$$AD = \sqrt{(14-x_D)^2 + (-7-(-10))^2}$$
$$AD = \sqrt{(14-x_D )^2 + (-7+10)^{2}}$$
Since $$D$$ lies on the y-axis, $$x_D = 0$$:
$$AD = \sqrt{(14)^2 + (3)^{2}} = \sqrt{205}$$
So, 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) Equation of the Perpendicular Bisector:
The equation of the perpendicular bisector of $$AB$$ can be found by first determining the midpoint of $$AB$$, which is the point $$M$$, and then finding the negative reciprocal of the slope of $$AB$$ to get the slope of the perpendicular bisector.
The coordinates of the midpoint $$M$$ are given by the average of the $$x$$-coordinates and the $$y$$-coordinates of $$A$$ and $$B$$:
$$M\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$$
Substituting in the coordinates of $$A(1, 1)$$ and $$B(5, 9)$$:
$$M\left(\frac{1 + 5}{2}, \frac{1 + 9}{2}\right) = (3, 5)$$
The slope of $$AB$$ is given by:
$$m_{AB} = \frac{y_B – y_A}{x_B – x_A}$$
Substituting in the coordinates of $$A$$ and $$B$$:
$$m_{AB} = \frac{9 – 1}{5 – 1} = \frac{8}{4} = 2$$
The slope of the perpendicular bisector ($$m_{\text{perpendicular}}$$) is the negative reciprocal of $$m_{AB}$$:
$$m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{1}{2}$$
Equation of the Perpendicular Bisector:
Using the point-slope form of a line $$(y – y_1) = m(x – x_1)$$, where $$(x_1, y_1)$$ is a point on the line, substitute $$M(3, 5)$$ and
(m_{\text{perpendicular}} = -\frac{1}{2}\):
$$y – 5 = -\frac{1}{2}(x – 3)$$
$$2y – 10 = 3 – x$$
$$2y = 13 – x$$
Therefore, the equation of the perpendicular bisector is $$2y = 13 – x$$.
(ii) Finding $$C$$ and $$D): The equation of the curve: \(6y = 5x^2 – 18x + 19$$
$$6\left(\frac{13 – x}{2}\right) = 5x^2 – 18x + 19$$
$$-3x + 39 = 5x^2 – 18x + 19$$
$$5x^2 – 15x – 20 = 0$$
$$5(x^2 – 3x – 4) = 0$$
$$5(x – 4)(x + 1) = 0$$
This gives two solutions for $$x$$: $$x = 4$$ or $$x = -1$$.
Now, substitute these values back into the equation of the perpendicular bisector to find the corresponding $$y$$-coordinates.
For $$x = 4$$:
$$2y = 13 – 4$$
$$2y = 9$$
$$y = 4.5$$
For $$x = -1$$:
$$2y = 13 – (-1)$$
$$2y = 14$$
$$y = 7$$
So, the points $$C$$ and $$D$$ are $$(4, 4.5)$$ and $$(-1, 7)$$, respectively.
The distance formula to find the distance $$CD$$:
$$CD = \sqrt{(x_D – x_C)^2 + (y_D – y_C)^2}$$
Substitute the coordinates of $$C$$ and $$D$$:
$$CD = \sqrt{(-1 – 4)^2 + (7 – 4.5)^2}$$
$$CD = \sqrt{(-5)^2 + (2.5)^2}$$
$$CD = \sqrt{25 + 6.25}$$
$$CD = \sqrt{\frac{125}{4}}$$
So, the distance $$CD$$ is $$\frac{\sqrt{125}}{2}$$.

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