The equation of the line with intercepts a and b is x/a + y/b = 1

Given: 

The line \(\frac{x}{a}+\frac{y}{b}=1\) intersects the axes (a,0) and (0,b). 

Explanation: 

So, (-4,3) divides the line segment AB and the ratio 5:3

-4 =   \(\frac{5+3a}{5+3},3=\frac{5b}{5+3}=1\) 

⇒ a = \(-\frac{32}{3},b=\frac{24}{5}\)

So, the equation of the line is   \(\frac{x}{\frac{32}{5}}+\frac{y}{\frac{24}{5}}=1\) 

⇒ 9x – 20y = -96 

Hence, the equation of line is 9x – 20y = -96

Correct Answer – x-y-8=0

Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0 

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0. 

(x + y – 3) + k(2x – y + 1) = 0 …..(i) 

x + y – 3 + 2kx – ky + k = 0

x + 2kx + y – ky – 3 + k = 0 

(1 + 2k)x + (1 – k)y – 3 + k = 0 

But, this line is parallel to X-axis Its slope = 0

⇒ \(\frac {-(1+2k)}{1-k} =0\)

⇒ 1 + 2k = 0

⇒ k = -1/2

Substituting the value of k in (i), we get

(x + y – 3) + -1/2 (2x – y + 1) = 0

⇒ 2(x + y – 3) – (2x – y + 1 ) = 0 

⇒ 2x + 2y – 6 – 2x + y – 1 = 0 

⇒ 3y – 7 = 0, which is the equation of the required line.

Correct Answer – 6x+11y-66=0 or x+2y-10=0
Let the required equation be `(x)/(a+5)+(y)/(a)=1`
Since it passes through (22,-6), we have `(22)/(a+b)-(6)/(a)=1`
This gives `a^(2)-11a+30=0 Rightarrow (a-6)(a-5)=0 Leftrightarrow a=6 or a=5`
So, the required equations are `(x)/(11)+(y)/(6)=1 or (x)/(10)+(x)/(5)=1`

Given, inclination of the line, θ = 30°.

Slope of the line, m = tan 30°

= \(\frac{1}{\sqrt{3}}\)

Equation of a line in one-point form is

y – y₀ = m(x – x₀)

Since m = \(\frac{1}{\sqrt{3}}\) and (x₀, y₀) = (2, 3)

∴ y − 3 = \(\frac{1}{\sqrt{3}}\)(x − 2)

⇒ \(\sqrt{3}\)y − 3\(\sqrt{3}\) = x – 2

⇒ x − \(\sqrt{3}\)y + (2 − 3\(\sqrt{3}\)) = 0, is the required equation of the line.

Correct Answer – `vec(r )=(hat(i)+2hat(j) +3hat(k)) + lambda (3hat(i) +2hat(j) -2hat(k))`
Clearly the required equation is `vec(r )=(hat(i) +2hat(j) +3hat(k)) + lambda(3hat(i) +2hat(j)-2hat(k))`

`vecr=veca+lambdavecb“vecr=(hati+2hatj+3hatk)+lambda(3hati+2hatj+2hatk)`