Answer:
a. Gradient of line AB is [tex]-\frac{1}{3}[/tex]
b. The gradient of a line perpendicular to line AB is 3
c. The equation of a line passing through point (4,2) and perpendicular to AB is [tex]y = 3x - 10[/tex]
Step-by-step explanation:
a.
Given
Point A (1, 3) B (7, 1)
Required
Gradient of AB
Gradient of a line is represented by m
m is calculated using the following formula
[tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}[/tex]
Where the general representation of the coordinates are [tex]A(x_1,y_1) and B(x_2,y_2)[/tex]
From the given data, we have that
[tex]A(x_1,y_1) = A(1,3)[/tex]
[tex]B(x_2,y_2) = A(7,1)[/tex]
So, from there we know that
[tex]x_1 = 1;y_1 =3; x_2 = 7;y_2 =1[/tex]
[tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}[/tex] becomes
[tex]m = \frac{1 - 3}{7 - 1}[/tex]
[tex]m = \frac{-2}{6}[/tex]
[tex]m = -\frac{1}{3}[/tex]
b.
Required
Find the gradient of a line perpendicular to AB
Recall that gradient of a line is represented by m
The condition for perpendicularity is that [tex]m_1.m_2 = -1[/tex]
In (a) above, we solved the gradient of line AB to be [tex]-\frac{1}{3}[/tex]
Let [tex]m_1[/tex] represent gradient of line AB
Hence, [tex]m_1 = -\frac{1}{3}[/tex]
Substitute [tex]-\frac{1}{3}[/tex] for [tex]m_1[/tex] in [tex]m_1.m_2 = -1[/tex]
This will give
[tex]\frac{-1}{3} * m_2 = -1[/tex]
Multiply both sides by -3
[tex]-3 * \frac{-1}{3} * m_2 = -1 * -3[/tex]
[tex]m_2 = -1 * -3[/tex]
[tex]m_2 = 3[/tex]
Hence, the gradient of a line perpendicular to line AB is 3
c.
Required
Find the equation of a line passing through point (4,2) and perpendicular to AB
Equation is calculated using the gradient formula
[tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}[/tex]
Since only one point is known, the formula is represented as follows
[tex]m = \frac{y - y_{1} }{x - x_{1}}[/tex]
Where [tex]x_1 = 4; y_1 = 2[/tex]
Since, the line is perpendicular to line AB, then its gradient m is equal to 3 (as calculated in b above)
So, we have [tex]x_1 = 4; y_1 = 2; m = 3[/tex]
By substitution
[tex]m = \frac{y - y_{1} }{x - x_{1}}[/tex] becomes
[tex]3 = \frac{y - 2}{x - 4}[/tex]
Multiply both sides by x - 4
[tex]3 * (x - 4) = \frac{y - 2}{x - 4} * (x - 4)[/tex]
[tex]3(x - 4) = {y - 2}[/tex]
Open brackets
[tex]3x - 12 = y - 2[/tex]
Make y the subject of formula
[tex]3x - 12 + 2= y[/tex]
[tex]3x - 10 = y[/tex]
Reorder
[tex]y = 3x - 10[/tex]
Hence, the equation of a line passing through point (4,2) and perpendicular to AB is [tex]y = 3x - 10[/tex]
