**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.**

**▶️Answer/Explanation**

(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**

(ii)Find the distance AD.

**▶️Answer/Explanation**

(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, \(y = 2x – 10\).

(ii) Distance AD:

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.**

**▶️Answer/Explanation**

(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 \(\sqrt{\frac{125}{4}}\).