A line L_1 is given by the equation 3x + 4y - 24 = 0, intersecting the x-axis at point A a… — JAMB UTME

A line $L_1$ is given by the equation $3x + 4y - 24 = 0$ , intersecting the x-axis at point $A$ and the y-axis at point $B$ . A second line $L_2$ has the equation $4x - 3y + k = 0$ , where $k > 0$ . If the perpendicular distance from the midpoint of the line segment $AB$ to $L_2$ is $5$ units, what is the value of $k$ ?

Answer: 18

Answer

The value of

k

18

To find

k

, first determine the intercepts of

L_1

: setting

y=0

gives

A(8, 0)

and setting

x=0

gives

B(0, 6)

. The midpoint

M

of segment

AB

\left(\frac{8+0}{2}, \frac{0+6}{2}\right) = (4, 3)

. Using the perpendicular distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

for point

M(4,3)

and line

4x - 3y + k = 0

, we get

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|7 + k|}{5}

. Setting

d = 5

gives

|7 + k| = 25

. Since

k > 0

, solving

7 + k = 25

gives the correct value

k = 18

Step-by-Step Solution

Determine the coordinates of points A and B

A = (8, 0)

and

B = (0, 6)

Setting

y = 0

3x + 4y - 24 = 0

yields

3x = 24 \implies x = 8

. Setting

x = 0

yields

4y = 24 \implies y = 6

Calculate the midpoint M of segment AB

M = (4, 3)

Using the midpoint formula

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{8 + 0}{2}, \frac{0 + 6}{2}\right) = (4, 3)

Set up the perpendicular distance equation from M(4,3) to line L₂

d = \frac{|7 + k|}{5}

Applying the distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

gives

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|16 - 9 + k|}{\sqrt{25}} = \frac{|7 + k|}{5}

Solve for k given d = 5 and k > 0

k = 18

Equating distance to

5

gives

\frac{|7 + k|}{5} = 5 \implies |7 + k| = 25

. Since

k > 0

7 + k = 25

, which yields

k = 18

Key Concept

Perpendicular Distance from a Point to a Straight Line

Estimated Time:2m 30s

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