Dear Student, Please find below the solution to the asked query: Let equilateral triangle be ABCx2+4xy+y2=0 ;iAbove equation represents pair of straight lines passing through origin, henceA0,0 be one of the vertices x-y=4⇒x=y+4Put value of x in iy+42+4y+4y+y2=0⇒y2+16+8y+4y2+16y+y2=0⇒6y2+24y+16=0⇒3y2+12y+8=0By Sridharacharya's formula, we get:y=-12±122-4×3×82×3=-12±486=-12±436=-2±233Hence y=-2+233 or y=-2-233When y=-2+233x=y+4=-2+233+4=2+233Hence coordinates of B will be B2+233,-2+233Side of equilateral trianglea=AB=2+233-02+-2+233-02=2+2332+-2+2332=22+2332+2 2 233+22+2332-2 2 233=4+43+4+43=8+83=323=423=aArea of equilateral triangle=34a2=344232=34×16×23=83According to ques83=A3Hence A=8 Answer Hope this information will clear your doubts about this topic If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible Regards

Question 3 pls

Question 3 pls 3. A family of lines is given by (1 + 2X)x + (1 — + X = O, X being a parameter. The line belonging to this family at the maximum distance from the point (1, 4) is ax + by + c = O, then the value of b—a is A point moves in the x-y plane such that the sum of its distances from *two mutually perpendicular lines is always equal to units. The area enclosed by the locus of the point is (in sq. units) The lines x2 + 4xy + Y2 = O and x — y = 4 are the sides of an equilateral triangle whose area is then A is equal to LINKED COMPREHENSION TYPE Read the following passage and answer the questions that follow. A triangle ABC is given where vertex A is (I, I) and the orthocenter is (2, 4). Also sides AB and BC are members of the family of lines ax + by + c = O where a, b, c are in A.P. Based on above written, solve the following questions. The vertex B is (D) none of these

Dear Student,
Please find below the solution to the asked query:

Let equilateral triangle be ABC x^{2} + 4 xy + y^{2} = 0; (i) Above equation represents pair of straight lines passing through origin, hence A (0, 0) be one of the vertices . x - y = 4 \Rightarrow x = y + 4 Put value of x in (i) {(y + 4)}^{2} + 4 (y + 4) y + y^{2} = 0 \Rightarrow y^{2} + 16 + 8 y + 4 y^{2} + 16 y + y^{2} = 0 \Rightarrow 6 y^{2} + 24 y + 16 = 0 \Rightarrow 3 y^{2} + 12 y + 8 = 0 By Sridharacharya' s formula, we get : y = \frac{- 12 \pm \sqrt{12^{2} - 4 \times 3 \times 8}}{2 \times 3} = \frac{- 12 \pm \sqrt{48}}{6} = \frac{- 12 \pm 4 \sqrt{3}}{6} = - 2 \pm \frac{2}{3} \sqrt{3} Hence y = - 2 + \frac{2}{3} \sqrt{3} or y = - 2 - \frac{2}{3} \sqrt{3} When y = - 2 + \frac{2}{3} \sqrt{3} x = y + 4 = - 2 + \frac{2}{3} \sqrt{3} + 4 = 2 + \frac{2}{3} \sqrt{3} Hence coordinates of B will be B (2 + \frac{2}{3} \sqrt{3}, - 2 + \frac{2}{3} \sqrt{3}) Side of equilateral triangle (a) = AB = \sqrt{{(2 + \frac{2}{3} \sqrt{3} - 0)}^{2} + {(- 2 + \frac{2}{3} \sqrt{3} - 0)}^{2}} = \sqrt{{(2 + \frac{2}{3} \sqrt{3})}^{2} + {(- 2 + \frac{2}{3} \sqrt{3})}^{2}} = \sqrt{2^{2} + {(\frac{2}{3} \sqrt{3})}^{2} + 2.2 . \frac{2}{3} \sqrt{3} + 2^{2} + {(\frac{2}{3} \sqrt{3})}^{2} - 2.2 . \frac{2}{3} \sqrt{3}} = \sqrt{4 + \frac{4}{3} + 4 + \frac{4}{3}} = \sqrt{8 + \frac{8}{3}} = \sqrt{\frac{32}{3}} = 4 \sqrt{\frac{2}{3}} = a Area of equilateral triangle = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} {(4 \sqrt{\frac{2}{3}})}^{2} = \frac{\sqrt{3}}{4} \times 16 \times \frac{2}{3} = \frac{8}{\sqrt{3}} According to ques \frac{8}{\sqrt{3}} = \frac{A}{\sqrt{3}} Hence A = 8 (Answer)

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