JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Jan Shift 1Standard
Question
A line passing through the point A(9,0)A(9, 0) makes an angle of 30°30° with the positive direction of the x-axis. If this line is rotated about A through an angle of 15°15° in the clockwise direction, then its equation in the new position is:
(A)y3−2+x=9\frac{y}{\sqrt{3}-2} + x = 9
(B)x3−2+y=9\frac{x}{\sqrt{3}-2} + y = 9
(C)x3+2+y=9\frac{x}{\sqrt{3}+2} + y = 9
(D)y3+2+x=9\frac{y}{\sqrt{3}+2} + x = 9
Solution Path
After 15∘15^\circ clockwise rotation, new angle =15∘= 15^\circ, m=2−3m = 2 - \sqrt{3}. Line: y3−2+x=9\frac{y}{\sqrt{3}-2} + x = 9.
01Question Setup
1/4
A line through A(9,0)A(9,0) at 30∘30^\circ to x-axis is rotated 15∘15^\circ clockwise about AA. Find the new equation.
30∘→15∘30^\circ \to 15^\circ rotation
02New Angle
2/4
New angle =30∘−15∘=15∘= 30^\circ - 15^\circ = 15^\circ. Slope m=tan⁡15∘=2−3m = \tan 15^\circ = 2 - \sqrt{3}.
m=tan⁡15∘=2−3m = \tan 15^\circ = 2 - \sqrt{3}
03Line EquationKEY INSIGHT
3/4
Point-slope form: y=(2−3)(x−9)y = (2-\sqrt{3})(x-9). Rearranging: y3−2+x=9\frac{y}{\sqrt{3}-2} + x = 9.
y3−2+x=9\frac{y}{\sqrt{3}-2} + x = 9
04Final Answer
4/4
The equation in the new position is y3−2+x=9\frac{y}{\sqrt{3}-2} + x = 9.
y3−2+x=9\boxed{\frac{y}{\sqrt{3}-2} + x = 9}
Concepts from this question1 concepts unlocked
★

Angle Between Two Lines

EASY

Given slopes m₁ and m₂, the tangent of the acute angle between the lines is |m₁-m₂|/(1+m₁m₂).

tan⁡θ=∣m1−m2∣1+m1m2\tan\theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}

Appears directly in 2-3 JEE questions every year. Quick formula application once you have the slopes.

Angle problemsBisector problemsTrisection
Practice (15 Qs) →
Want more practice?
Try more PYQs from this chapter →