Maths-

General

Easy

Question

# Find a value for k so that the line through (k, 6) and (4, 2) is perpendicular to the line with equation x + 2y = 2.Application

Hint:

### 1. Slopes of perpendicular lines are negative reciprocals of each other.

2. When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

## The correct answer is: Value of k for the given line is 6.

### Step-by-step solution:-

x + 2y = 2

∴ 2y = -x + 2

∴ y = -1/2 x + 1 ................................................ (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -1/2 ......................... (Equation i)

The given line is perpendicular to x + 2y = 2 and we know that perpendicular lines have slopes that are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line (x + 2y = 2)

∴ Slope of the given line = m = -1/ -1/2 = 2 ..................... (From Equation i) .................................................................... (Equation ii)

The given line passes through the point (k,6) & (4,2).

Hence, x1 = k, y1 = 6, x2 = 4 & y2 = 2

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

∴ (y-6) = (2-6) × (x-k)

4-k

∴ y - 6 = -4 × (x - k)

4-k

∴ (y-6) × (4-k) = -4 ×(x-k)

∴ 4y - ky - 24 + 6k = -4x + 4k

∴ 4y - ky = -4x + 4k - 6k + 24

∴ (4-k) y = -4x -2k + 24

∴ y = -4/(4-k) x - 2/(4-k) k + 24/(4-k) ................................................. (Equation iii)

Comparing Equation iii with the standard form of a straight line i.e. y = mx + c, we get-

m = - 4/(4-k)

∴ 2 = - 4/(4-k) ..................................... (From Equation ii)

∴ 2 (4-k) = - 4 ...................................... [(Multiplying both sides by (4-k)]

∴ 8 - 2k = - 4

∴ -2k = - 4 - 8

∴ -2k = -12

∴ k = 6 .............................................. (Dividing both sides by -2)

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