Question 53 of 70intermediate🔧 ApplyShort Answer4 marks

Find the equation of a line drawn perpendicular to the line x4+y6=1\frac{x}{4} + \frac{y}{6} = 1 through the point, where it meets the yy-axis.

Correct Answer

The equation of the required line is 2x−3y+18=02x - 3y + 18 = 0 or y=23x+6y = \frac{2}{3}x + 6.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 6 | (Page 22)
For More Understanding

Explanation

This question tests understanding of intercept form of a line, finding y-intercept, perpendicular lines, and point-slope form. The student needs to: (1) identify the y-intercept from the intercept form, (2) find the slope of the given line, (3) use the perpendicular condition m₁m₂ = -1, and (4) apply point-slope form to get the final equation.

Solution Steps

  1. Step 1: The given line is x4+y6=1\frac{x}{4} + \frac{y}{6} = 1. To find where it meets the y-axis, put x = 0. We get y6=1\frac{y}{6} = 1, so y = 6. The point is (0, 6).

  2. Step 2: Convert to slope-intercept form: x4+y6=1\frac{x}{4} + \frac{y}{6} = 1 becomes 3x+2y=123x + 2y = 12, which gives y=−32x+6y = -\frac{3}{2}x + 6. So the slope of given line is m1=−32m_1 = -\frac{3}{2}.

  3. Step 3: For perpendicular lines, m1×m2=−1m_1 \times m_2 = -1. So −32×m2=−1-\frac{3}{2} \times m_2 = -1, giving m2=23m_2 = \frac{2}{3}.

  4. Step 4: Using point-slope form y−y0=m(x−x0)y - y_0 = m(x - x_0) for the perpendicular line through (0, 6) with slope 23\frac{2}{3}: y−6=23(x−0)y - 6 = \frac{2}{3}(x - 0), which simplifies to 2x−3y+18=02x - 3y + 18 = 0.

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