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) M=(1,4) \(gradient=\frac{1}{2}\)
Gradient of MB=-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
(ii)Find the distance AD.
▶️Answer/Explanation
(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.
▶️Answer/Explanation
(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 \(\frac{\sqrt{125}}{2}\).