Question

The line segment joining P (5,-2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A on it Find the equation of a line through the point A perpendicular on the line x-3y +4-0.​

Answer 1

Step-by-step explanation:

Given that:The line segment joining P (5,-2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A on it Find the equation of a line through the point A perpendicular on the line x-3y +4-0.

To find:Find the equation of a line through the point A perpendicular on the line x-3y +4-0.

Solution:

To find line passing through A,

first find the coordinates of point A by applying section formula

Section formula:

Let Point A(x1,y1) and B(x2,y2) are divided by point C(x,y) in ratio m:n,then coordinates of C are

$\boxed{\red{x = \frac{mx_2 + nx_1}{m + n}}} \\ \\\boxed{\red{ y = \frac{my_2 + ny_1}{m + n}}}$

To find A:The line segment joining P (5,-2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A

$x = \frac{3 \times 9 + 1 \times 5}{3 + 1} \\ \\ x = \frac{27 + 5}{4} \\ \\ x = \frac{32}{4} \\ \\\bold{\red{x = 8}} \\ \\ y = \frac{3 \times 6 + 1 \times ( - 2)}{3 + 1} \\ \\y = \frac{18 - 2}{4} \\ \\ y = \frac{16}{4} \\ \\ \bold{\red{y= 4}}$

Coordinates of A(8,4)

To find line passing through a point

$\boxed{\bold{\blue{y - y_1 = m(x - x_1) }}}$

To find slope: Use

$x-3y +4 = 0$

Convert this line in slope intercepted form,i. e. y=mx+c

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

Solve of line x-3y+4=0 is 1/3

Solpe of line perpendicular to x-3y+4=0 is -3.

Solpe of two lines ,if they are perpendicular

$\boxed{\bold{m_1m_2 = - 1 }}$

Thus,equation of line passing through A(8,4) having slope -3

$y - 4 = - 3(x - 8) \\ \\ y - 4= - 3x + 24 \\ \\ 3x + y = 24 + 4 \\ \\ \bold{\green{3x + y = 28 }}$

Hope it helps you.

Answer 2

Step-by-step explanation:

hope it helpd u

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