Question

The line on the graph passes through to points A (1, 3) B (7, 1)

Answer 1

Answer:

a. Gradient of line AB is $-\frac{1}{3}$

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 $y = 3x - 10$

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

$m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}$

Where the general representation of the coordinates are $A(x_1,y_1) and B(x_2,y_2)$

From the given data, we have that

$A(x_1,y_1) = A(1,3)$

$B(x_2,y_2) = A(7,1)$

So, from there we know that

$x_1 = 1;y_1 =3; x_2 = 7;y_2 =1$

$m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}$ becomes

$m = \frac{1 - 3}{7 - 1}$

$m = \frac{-2}{6}$

$m = -\frac{1}{3}$

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 $m_1.m_2 = -1$

In (a) above, we solved the gradient of line AB to be $-\frac{1}{3}$

Let $m_1$ represent gradient of line AB

Hence, $m_1 = -\frac{1}{3}$

Substitute $-\frac{1}{3}$ for $m_1$ in  $m_1.m_2 = -1$

This will give

$\frac{-1}{3} * m_2 = -1$

Multiply both sides by -3

$-3 * \frac{-1}{3} * m_2 = -1 * -3$

$m_2 = -1 * -3$

$m_2 = 3$

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

$m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}$

Since only one point is known, the formula is represented as follows

$m = \frac{y - y_{1} }{x - x_{1}}$

Where $x_1 = 4; y_1 = 2$

Since, the line is perpendicular to line AB, then its gradient m is equal to 3 (as calculated in b above)

So, we have $x_1 = 4; y_1 = 2; m = 3$

By substitution

$m = \frac{y - y_{1} }{x - x_{1}}$ becomes

$3 = \frac{y - 2}{x - 4}$

Multiply both sides by x - 4

$3 * (x - 4) = \frac{y - 2}{x - 4} * (x - 4)$

$3(x - 4) = {y - 2}$

Open brackets

$3x - 12 = y - 2$

Make y the subject of formula

$3x - 12 + 2= y$

$3x - 10 = y$

Reorder

$y = 3x - 10$

Hence, the equation of a line passing through point (4,2) and perpendicular to AB is $y = 3x - 10$