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NCERT Exemplar · Q22

Q.A line cutting off intercept −3-3 from the yy-axis and the tangent at angle to the xx-axis is 35\dfrac{3}{5}, its equation is
(A) 5y−3x+15=05y-3x+15=0
(B) 3y−5x+15=03y-5x+15=0
(C) 5y−3x−15=05y-3x-15=0
(D) None of these

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A line with yy-intercept −3-3 and slope m=35m = \frac{3}{5} (since tangent of the angle with the xx-axis equals the slope) has equation y=35x−3y = \frac{3}{5}x - 3, which rearranges to 5y−3x+15=05y - 3x + 15 = 0.

The phrase "tangent at angle to the xx-axis" means the tangent of the angle the line makes with the positive xx-axis. This tangent value is precisely the slope of the line. So when we're told the tangent is 35\frac{3}{5}, we immediately know m=35m = \frac{3}{5}.

The intercept "cut off from the yy-axis" is the yy-coordinate where the line crosses the yy-axis, which is the yy-intercept cc. Here it's given as −3-3.

With slope and yy-intercept in hand, we can write the equation in slope-intercept form and then convert to standard form.

Step-by-step construction

  1. Identify the slope. The tangent of the angle with the xx-axis is 35\frac{3}{5}, so the slope is:

m=35m = \frac{3}{5}

  1. Identify the yy-intercept. The line cuts off an intercept of −3-3 from the yy-axis, meaning it crosses at (0,−3)(0, -3). Thus:

c=−3c = -3

  1. Write the slope-intercept form. Using y=mx+cy = mx + c: …

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