Question

Find the equation of line which passes through the point (22,-6) and whose intercept on the x-axis exceed the intersecp on y-axis by 5.

Answer 1

let intercet on y axis is a
thus intercept on x axis is a+ 5
equation is x/a +y/a+5=1
xa=a square +5a-ay-5y mark it equation no. 1
slope
of line is -a/a+5
equation in point slope form is
y+6=-a/a+5 × (x-22)
€ ay+ax+5y-16a+30=0 mark this as equation no. 2
put equation 1 in 2nd we get
a square -11a+30 =0
and a=5,6

Answer 2

The equation of line which passes through the point (22,-6) is, x+2y−10 and 6x+11y−66=0.

What is a equation of the line?

• The standard form of equation of a line is ax + by + c = 0.
• Here a, b, are the coefficients, x, y are the variables, and c is the constant term.
• It is an equation of degree one, with variables x and y.
• The values of x and y represent the coordinates of the point on the line represented in the coordinate plane.

According to the question:

Let equation of line be $\frac{x}{a}+\frac{y}{b}=1$.

Here, y− intercept is b.

        x− intercept is a.

Also again a=b+5.

So  $\frac{x}{b+5}+\frac{y}{b}=1$ , it pass through (22,−6).

$\frac{22}{b+5}-\frac{6}{b}=1 \\\\\Rightarrow 22 b-b(b+5)\\\\=6(b+5)$

$b^{2}+5 b=22 b-6 b-30\\b^{2}+5 b=16 b-30\\ b^{2}-11 b+30=0 \Rightarrow(b-5)(b-6)=0$.

b=5,6.

so, a=10 for b=5.

    a=11 for b=6.

equation of lines are $\frac{x}{10}+\frac{y}{5}=1$ .

$\frac{x}{11}+\frac{y}{6}=1$.

x+2y=10 and 6x+11y=66.

x+2y−10 and 6x+11y−66=0.

Hence, The equation of line which passes through the point (22,-6) is, x+2y−10 and 6x+11y−66=0.

Learn more about equation of line here,

https://brainly.in/question/48103092?msp_poc_exp=5

#SPJ2