CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard

Class 10
Solutions of Sample Papers for Class 10 Boards

## (iii)What are the coordinates of the point on y axis equidistant from A and G?

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### Transcript

Question 36 (iii) β Choice 1 What are the coordinates of the point on y axis equidistant from A and G? Here, A(-2,2) and G(-4,7) Since the required point is in y-axis, its x βcoordinate will be zero Let Required point = Z (0, a) As per question, point Z is equidistant from A & B Hence, AZ = BZ Finding AC x1 = β2 , y1 = 2 x2 = 0 , y2 = a AC = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β(( 0 β(β2))2+(πβ2)2) = β((2)2+(πβ2)2) = β((2)2+ π2+22β2(2)(π) ) = β(4+ π2+4β4π) = β(ππβππ+π) Finding BC x1 = β4 , y1 = 7 x2 = 0 , y2 = a BC = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β(( 0 β(β4))2+(7 βπ)2) = β((0+4)2+(7 βπ)2) = β((4)2+(7 βπ)2) = β((4)2+ π2+72β2(7)(π) ) = β(16+ π2+49β14π) = β(ππβπππ+ππ) Now, AC = BC β(π2β4π+8) = β(π2β14π+65) Squaring both sides (β(π2β4π+8) " )2 = (" β(π2β14π+65))^2 ππβππ+π = ππβπππ+ππ β4a + 8 = β14a + 65 β4a + 14a = 65 β 8 10a = 57 a = 57/10 a = 5.7 Hence the required point is C(0, a) = (0, 9)