CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard
You are here
CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard
Last updated at Aug. 26, 2025 by Teachoo
Get access to this page and all premium solutions, videos, and worksheets by joining Teachoo Black.
Transcript
Question 37 (iii) - Choice (A) Find the point on x-axis which is equidistant from P and Q.Since the point is on the x−axis, its y-coordinate will be zero ∴ y = 0 Let the required point be M (a, 0) Given that Point M is equidistant from P & Q Hence, PM = QM Finding PM x1 = 2, y1 = 5 x2 = a, y2 = 0 PM = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −2)2+(0−5)2) = √((𝑎−2)^2+(−5)2) = √(𝑎2+22 −2(2)(𝑎)+(5)2) = √(𝑎2+4 −4𝑎+25) = √(𝒂𝟐 −𝟒𝒂+𝟐𝟗) Finding QM x1 = 4, y1 = 4 x2 = a, y2 = 0 QM = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −4)2+(0−4)2) = √((𝑎−4)^2+(−4)2) = √(𝑎2+42−2(4)(𝑎)+(4)2) = √(𝑎2+16−8𝑎+16) = √(𝒂𝟐−𝟖𝒂+𝟑𝟐) Now, PM = QM √(𝒂𝟐 −𝟒𝒂+𝟐𝟗) = √(𝒂𝟐−𝟖𝒂+𝟑𝟐) Squaring both sides (√(𝑎2 −4𝑎+29) )2 = (√(𝑎2−8𝑎+32))2 a2 – 4a + 29 = a2 – 8a + 32 a2 − a2 + 8a – 4a = 32 – 29 4a = 3 a = 𝟑/𝟒 Hence the required point is M(a, 0) i.e. (𝟑/𝟒, 0)
