Distance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Given: P(-3, 8) and Q(5, -7) Here, x₁ = -3, y₁ = 8, x₂ = 5, y₂ = -7 PQ = √[(5-(-3))² + (-7-8)²] = √[(5+3)² + (-15)²] = √[8² + (-15)²] = √[64 + 225] = √289 = 17 units ∴ Distance PQ = 17 unitsDistance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Given: A(-2, 3), B(4, 3), C(4, -5) Finding all three distances: Distance AB: AB = √[(4-(-2))² + (3-3)²] = √[6² + 0²] = √36 = 6 units Distance BC: BC = √[(4-4)² + (-5-3)²] = √[0² + (-8)²] = √64 = 8 units Distance AC: AC = √[(4-(-2))² + (-5-3)²] = √[6² + (-8)²] = √[36 + 64] = √100 = 10 units Comparing: AB < BC < AC 6 < 8 < 10 ∴ Closest villages: A and B (6 units) ∴ Farthest villages: A and C (10 units)Midpoint Formula: M = ((x₁+x₂)/2, (y₁+y₂)/2) Given: R(-8, 11) and S(4, -5) Here, x₁ = -8, y₁ = 11, x₂ = 4, y₂ = -5 M = ((-8+4)/2, (11+(-5))/2) = ((-4)/2, (6)/2) = (-2, 3) ∴ Midpoint M = (-2, 3)Midpoint Formula: M = ((x₁+x₂)/2, (y₁+y₂)/2) Given: P(2a, 4), Q(-4, 5b), M(3, 7) For x-coordinate: (2a + (-4))/2 = 3 2a - 4 = 6 2a = 10 a = 5 For y-coordinate: (4 + 5b)/2 = 7 4 + 5b = 14 5b = 10 b = 2 Verification: Midpoint = ((2×5-4)/2, (4+5×2)/2) = ((10-4)/2, (4+10)/2) = (6/2, 14/2) = (3, 7) ∴ a = 5 and b = 2Midpoint Formula: M = ((x₁+x₂)/2, (y₁+y₂)/2) Given: M(5, -2), Q(-1, 8) Let P = (x, y) For x-coordinate: (x + (-1))/2 = 5 (x - 1)/2 = 5 x - 1 = 10 x = 11 For y-coordinate: (y + 8)/2 = -2 y + 8 = -4 y = -12 Verification: Midpoint of P(11,-12) and Q(-1,8) = ((11-1)/2, (-12+8)/2) = (10/2, -4/2) = (5, -2) ∴ P = (11, -12)Section Formula (for ratio m:n): P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) Here instead of m:n user can take m1:m2 Given: A(-6, 12) and B(9, -6) Trisection means dividing into 3 equal parts First trisection point P divides AB in ratio 1:2 P = ((1×9 + 2×(-6))/(1+2), (1×(-6) + 2×12)/(1+2)) = ((9-12)/3, (-6+24)/3) = (-3/3, 18/3) = (-1, 6) Second trisection point Q divides AB in ratio 2:1 Q = ((2×9 + 1×(-6))/(2+1), (2×(-6) + 1×12)/(2+1)) = ((18-6)/3, (-12+12)/3) = (12/3, 0/3) = (4, 0) ∴ Trisection points are P(-1, 6) and Q(4, 0) Here instead of P and Q user can use any alphabet other then A and B Here user has to draw a line with point A and B at the end.The line should have two points P and Q where P is closer to A and Q is closer to B.Section Formula (for ratio m:n): Point = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) Here instead of m:n user can take m1:m2 Given: P(5, -4) and Q(-7, 8) X is closer to P, so X divides PQ in ratio 1:2 Finding X: X = ((1×(-7) + 2×5)/(1+2), (1×8 + 2×(-4))/(1+2)) = ((-7+10)/3, (8-8)/3) = (3/3, 0/3) = (1, 0) Y divides PQ in ratio 2:1 (closer to Q) Y = ((2×(-7) + 1×5)/(2+1), (2×8 + 1×(-4))/(2+1)) = ((-14+5)/3, (16-4)/3) = (-9/3, 12/3) = (-3, 4) ∴ X = (1, 0) and Y = (-3, 4) Here user has to draw a line with point P and Q at the end.The line should have two points X and Y where X is closer to P and Y is closer to Q.Section Formula (for internal division in ratio m:n): P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)) Here instead of m:n user can take m1:m2 Given: A(6, -3), B(-4, 12), ratio = 2:3 Here, x₁ = 6, y₁ = -3, x₂ = -4, y₂ = 12, m = 2, n = 3 P = ((2×(-4) + 3×6)/(2+3), (2×12 + 3×(-3))/(2+3)) = ((-8+18)/5, (24-9)/5) = (10/5, 15/5) = (2, 3) ∴ Point of division = (2, 3)Section Formula (for ratio k:1): P = ((kx₂ + 1×x₁)/(k+1), (ky₂ + 1×y₁)/(k+1)) Given: A(5, -2), B(-8, 11), P(-1, 4) Let the ratio be k:1 (-1, 4) = ((k×(-8) + 1×5)/(k+1), (k×11 + 1×(-2))/(k+1)) From x-coordinate: -1 = (-8k + 5)/(k+1) -1(k+1) = -8k + 5 -k - 1 = -8k + 5 7k = 6 k = 6/7 Verification with y-coordinate: 4 = (11k - 2)/(k+1) 4(k+1) = 11k - 2 4k + 4 = 11k - 2 6 = 7k k = 6/7 ∴ Ratio = 6/7 : 1 = 6:7Section Formula (for ratio m:n): y-coordinate = (m×y₂ + n×y₁)/(m+n) Here instead of m:n user can take m1:m2 Given: A(6, 9), B(-4, -6) On x-axis, y-coordinate = 0 Let the ratio be m:n 0 = (m×(-6) + n×9)/(m+n) 0 = -6m + 9n 6m = 9n m/n = 9/6 = 3/2 ∴ Ratio = 3:2 Finding x-coordinate of point of division: x = (3×(-4) + 2×6)/(3+2) = (-12 + 12)/5 = 0/5 = 0 ∴ Point of division = (0, 0) ∴ Ratio is 3:2 and point is (0, 0)Midpoint Formula: M = ((x₁+x₂)/2, (y₁+y₂)/2) Given: P(-3, 1), Q(5, 7), R(9, -3) Midpoint of PQ: M₁ = ((-3+5)/2, (1+7)/2) = (2/2, 8/2) = (1, 4) Midpoint of QR: M₂ = ((5+9)/2, (7+(-3))/2) = (14/2, 4/2) = (7, 2) Midpoint of RP: M₃ = ((9+(-3))/2, (-3+1)/2) = (6/2, -2/2) = (3, -1) ∴ Midpoints are: M₁(1, 4) - midpoint of PQ M₂(7, 2) - midpoint of QR M₃(3, -1) - midpoint of RPDistance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Given: Centre C(4, -5), P(12, -1) lies on circle Step 1: Find radius of circle Radius = CP = √[(12-4)² + (-1-(-5))²] = √[8² + 4²] = √[64 + 16] = √80 = 4√5 units Step 2: Find point Q on y-axis Q is on y-axis, so Q = (0, y) Since Q is also on the circle: CQ = CP √[(0-4)² + (y-(-5))²] = √80 √[16 + (y+5)²] = √80 16 + (y+5)² = 80 (y+5)² = 64 y + 5 = ±8 y = -5 + 8 = 3 or y = -5 - 8 = -13 Verification: For Q(0,3): CQ = √[16 + 64] = √80 For Q(0,-13): CQ = √[16 + 64] = √80 ∴ Q = (0, 3) or Q = (0, -13)Section Formula (for ratio m:n): Point = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) Given: C(0, 9), F(8, 3) Let D(x, y) F is the second trisection point (closer to D) So F divides CD in ratio 2:1 F = ((2x + 1×0)/(2+1), (2y + 1×9)/(2+1)) (8, 3) = ((2x)/3, (2y + 9)/3) From x-coordinate: 8 = 2x/3 24 = 2x x = 12 From y-coordinate: 3 = (2y + 9)/3 9 = 2y + 9 2y = 0 y = 0 ∴ D = (12, 0) Now finding E (first trisection point, ratio 1:2): E = ((1×12 + 2×0)/(1+2), (1×0 + 2×9)/(1+2)) = (12/3, 18/3) = (4, 6) ∴ D = (12, 0) and E = (4, 6) Here user has to draw the line segment CD and mark the points E and F where E should be closer to C and F should be closer to D.Section Formula (for ratio m:n): Point = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) Distance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Given: P(0, 0), Q(40, 0) (a) Finding R (ratio 3:5): R = ((3×40 + 5×0)/(3+5), (3×0 + 5×0)/(3+5)) = (120/8, 0/8) = (15, 0) Finding S (ratio 5:3): S = ((5×40 + 3×0)/(5+3), (5×0 + 3×0)/(5+3)) = (200/8, 0/8) = (25, 0) ∴ R = (15, 0) and S = (25, 0) (b) Distance RS: RS = √[(25-15)² + (0-0)²] = √[10² + 0] = √100 = 10 units (c) Distance PQ: PQ = √[(40-0)² + (0-0)²] = 40 units Fraction = RS/PQ = 10/40 = 1/4 ∴ (a) R = (15, 0), S = (25, 0) ∴ (b) RS = 10 units ∴ (c) RS = 1/4 of PQSection Formula (for ratio m:n): Point = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) Distance Formula: d = √[(x₂-x₁)² + (y₂-y₁)²] Given: A(2, 5), B(14, 17) (a) T is the second trisection point (closer to B) T divides AB in ratio 2:1 T = ((2×14 + 1×2)/(2+1), (2×17 + 1×5)/(2+1)) = ((28+2)/3, (34+5)/3) = (30/3, 39/3) = (10, 13) ∴ T = (10, 13) (b) "Same distance beyond T in opposite direction" Vector AT = T - A = (10-2, 13-5) = (8, 8) Distance AT = √[64+64] = √128 = 8√2 units Point C is such that we extend from T by same distance C = T + AT (extending beyond T) C = (10, 13) + (8, 8) = (18, 21) Verification: TC = √[(18-10)² + (21-13)²] = √[64+64] = 8√2 AT = TC (equal distances) ∴ (a) T = (10, 13) ∴ (b) C = (18, 21)
