For what values of a and b the intercepts cut off on the coordinate ax

1.	For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.
Answer» Given: Intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 ……(i) And are equal in length but opposite in sign to those cut off by the line 2x – 3y + 6 = 0 ……(ii) Explanation: The slope of two lines are equal The slope of the line (i) is \(-\frac{a}{b}\) The slope of the line (ii) is \(\frac{2}{3}\) ∴ \(-\frac{a}{b}=\frac{2}{3}\) a = \(-\frac{2b}{3}\) The length of the perpendicular from the origin to the line (i) is The formula used: d = \(\|\frac{ax+by+d}{\sqrt{a^2+b^2}}\|\) d₁ = \(\|\frac{a(0)+b(0)+8}{\sqrt{a^2+b^2}}\|\) d₁ = \(\frac{8\times3}{\sqrt{13b^2}}\) The length of the perpendicular from the origin to the line (ii) is The formula used: d = \(\|\frac{ax+by+d}{\sqrt{a^2+b^2}}\|\) d₂ = \(\|\frac{2(0)-3(0)+6}{\sqrt{2^2+3^2}}\|\) Given: d₁= d₂ \(\frac{8\times3}{\sqrt{13b^2}}\) = \(\frac{6}{\sqrt{13b}}\) ⇒ b = 4 ∴ a = \(\frac{2b}{3}\) = \(-\frac{8}{3}\) Hence the value of a and b is \(-\frac{8}{3}\) , 4.

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

Answer»

Given:

Intercepts cut off on the coordinate axes by the line

ax + by + 8 = 0 ……(i)

And are equal in length but opposite in sign to those cut off by the line

2x – 3y + 6 = 0 ……(ii)

Explanation:

The slope of two lines are equal

The slope of the line (i) is \(-\frac{a}{b}\)

The slope of the line (ii) is \(\frac{2}{3}\)

∴ \(-\frac{a}{b}=\frac{2}{3}\)

a = \(-\frac{2b}{3}\)

The length of the perpendicular from the origin to the line (i) is

The formula used: d = \(|\frac{ax+by+d}{\sqrt{a^2+b^2}}|\)

d₁ = \(|\frac{a(0)+b(0)+8}{\sqrt{a^2+b^2}}|\)

d₁ = \(\frac{8\times3}{\sqrt{13b^2}}\)

The length of the perpendicular from the origin to the line (ii) is

The formula used: d = \(|\frac{ax+by+d}{\sqrt{a^2+b^2}}|\)

d₂ = \(|\frac{2(0)-3(0)+6}{\sqrt{2^2+3^2}}|\)

Given: d₁= d₂

\(\frac{8\times3}{\sqrt{13b^2}}\) = \(\frac{6}{\sqrt{13b}}\)

⇒ b = 4

∴ a = \(\frac{2b}{3}\) = \(-\frac{8}{3}\)

Hence the value of a and b is \(-\frac{8}{3}\) , 4.