Question

Point R divides the line segment joining the points A(4,2)and B(4,-7)suchandigarh that AC/AB =1/3.if C lies on the line 3x-4y+k=0,the value of k

Answer 1

Answer:

k = -16

Step-by-step explanation:

Hi,

Given A (4, 2) and B(4, -7)

Given that C divided the line segment joining the points A and B

such that AC/AB = 1/3

⇒3AC = AB = AC + CB

⇒2AC = CB

⇒AC/CB = 1/2

Thus, C is point on AB which divided it in the ration 1: 2

Now, using internal sectional formula to find the coordinates of

the point C,

we get C(4 + 2.4/1+2, -7 + 2.2/1+2)

= C(12/3, -3/3)

= C(4, -1).

Now, given that C (4, -1) lies on the line 3x - 4y + k = 0

⇒3(4) -4(-1) + k = 0

⇒12 + 4 + k = 0

k = -16.

Hope, it helped !

Answer 2

Solution:

It is given that

AC:AB = 1:3

AC+BC=AB

AC+BC=3AC

3AC-AC=BC

2AC=BC

AC/BC=1/2

AC:BC = 1:2

by this way the point C internally divided the segment into 1:2

by section formula coordinates of C are

$x = \frac{8 + 4}{3} = \frac{12}{3} = 4 \\ \\ y = \frac{ - 7 + 4}{3} = \frac{ - 3}{3} = - 1$

Thus C(4,-1)

if C lies on the line 3x-4y+k=0

So

$3(4) - 4( - 1) + k = 0 \\ \\ 12 + 4 + k = 0 \\ \\ k = - 16$


Hope it helps you.