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A line perpendicular to 3x + 4y = 5 and passing through a point (4, 6) cuts a line 2x + 3y = 8 at a point:1. (1, 2)2. (1, 3)3. (2, 3)4. (0, 3)

Answer» Correct Answer - Option 1 : (1, 2)

Concept:

The general equation of a line is y = mx + c 

Where m is the slope and c is any constant

  • The slope of parallel lines is equal.
  • Slope of the perpendicular line have their product = -1
 

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

 

Calculation:

Given line 3x + 4y = 5

⇒ y = \(-3\over4\)x + \(5\over4\)

⇒ Slope(m1) = \(-3\over4\) and c1 = \(5\over4\)

Now for the slope of the perpendicular line (m2)

m1 × m2 = -1

⇒ \(-3\over4\) × m2 = -1 

⇒ m2 = \(4\over3\)

Perpendicular line has the slope \(4\over3\) and passes through (4, 6)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 6 = \(4\over3\) (x - 4)

⇒ 3y - 4x = 2

The intersection with the line 2x + 3y = 8 is:

(Substracting the 2 equations)

⇒ 6x = 6

⇒ x = 1

Putting value of x in equation of perpendicular line

⇒ y = 2

∴ The point if intersection is (1, 2)


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