Question

\( Y \)-अक्ष एवं रेखा \( 2 x+3 y=4 \) के प्रतिच्छेद-बिन्दु से हैंच जाने वाली उस रेखा का समीकरण ज्ञात कीजिए जो \( X \)-अक्ष से \( 60^{\circ} \) माक् कोण बनाती है।

Answer 1

Equation of y-axis is x = 0

line is 2x + 3y = 4

For their intersection, put x = 0 in 2x + 3y = 4

⇒ \(y =\frac 43\)

Hence, intersection point is \((0, \frac 43)\).

Angle of required line with +ve direction of x-axis is 60°.

\(\therefore \) Slope of required line is m = tan60° = √3.

Let required line is y = mx + C

⇒ \(y = \sqrt3 x + C\)     \((\because m = \sqrt 3)\)

⇒ \(\frac 43 = \sqrt 3 \times 0 + C\)    (By putting x = 0 & y = \(\frac 43\))

⇒ \(C = \frac 43\)

\(\therefore\) Equation of required line is \(y = \sqrt 3 x + \frac 43\).

Comments

Right