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LearnQuantitative AptitudeGeometry & Mensuration

04.06 · Geometry & Mensuration

Coordinate Geometry: Points & Lines

Elite lesson12 formulas10 questions

On the syllabus of

Elite lesson1 h 25 minHinglish

Formulas

At a glance 12 formulas

  1. 1Distance between two pointsd=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}points (x1,y1) and (x2,y2)\text{points } (x_1, y_1) \text{ and } (x_2, y_2)
  2. 2Section formulaP=(mx2+nx1m+n, my2+ny1m+n)P = \left(\dfrac{mx_2 + nx_1}{m + n},\ \dfrac{my_2 + ny_1}{m + n}\right)P splits (x1,y1) to (x2,y2) in m:nP \text{ splits } (x_1, y_1) \text{ to } (x_2, y_2) \text{ in } m : n
  3. 3Midpoint M, centroid GG=(x1+x2+x33, y1+y2+y33)G = \left(\tfrac{x_1 + x_2 + x_3}{3},\ \tfrac{y_1 + y_2 + y_3}{3}\right)M=(x1+x22,y1+y22); midpoints: same GM = \left(\tfrac{x_1 + x_2}{2}, \tfrac{y_1 + y_2}{2}\right);\ \text{midpoints: same } G
  4. 4Triangle from its corners12∣ad−bc∣ for (0,0),(a,b),(c,d)\tfrac{1}{2}|ad - bc|\ \text{for } (0, 0), (a, b), (c, d)shift any corner to (0,0) first\text{shift any corner to } (0, 0) \text{ first}
  5. 5Slopem=y2−y1x2−x1=tan⁡θm = \dfrac{y_2 - y_1}{x_2 - x_1} = \tan\thetaθ with x-axis; ax+by+c=0: m=−ab\theta \text{ with } x\text{-axis};\ ax + by + c = 0:\ m = -\tfrac{a}{b}
  6. 6Equation of a liney−y1=m(x−x1)y - y_1 = m(x - x_1)intercepts a and b: xa+yb=1\text{intercepts } a \text{ and } b:\ \dfrac{x}{a} + \dfrac{y}{b} = 1
  7. 7Parallel and perpendicular∥: m1=m2; ⊥: m1m2=−1\parallel:\ m_1 = m_2;\ \perp:\ m_1 m_2 = -1⊥ to ax+by=c: bx−ay=k\perp \text{ to } ax + by = c:\ bx - ay = k
  8. 8Angle between two linestan⁡θ=∣m1−m21+m1m2∣\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|θ acute; 1+m1m2=0⇒θ=90∘\theta \text{ acute};\ 1 + m_1 m_2 = 0 \Rightarrow \theta = 90^\circ
  9. 9Distance to a lined=∣ax1+by1+c∣a2+b2d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}parallel, same a,b: ∣c1−c2∣a2+b2\text{parallel, same } a, b:\ \dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}
  10. 10Three lines, no triangleThree lines fail to make a triangle if two are parallel (equal slopes) or all three pass through one point: solve two of them and put that point into the third.
  11. 11Smallest cut-off triangleA line through (h,k)(h, k), with h,k>0h, k > 0, cuts off at least area 2hk2hk with the positive axes; the least comes when (h,k)(h, k) is the midpoint of the cut-off part.
  12. 12Shortest path by reflectionTo go from A to a line and on to B (same side), reflect A in the line and join the image to B. In the x-axis (x,y)(x, y) goes to (x,−y)(x, -y); in y=xy = x, swap the coordinates.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    What is the slope of the line through (−1,−5)(-1, -5) and (3,7)(3, 7)?

    Correct answer33

    Why

    Slope is the change in yy over the change in xx, both taken in the same order.

    7−(−5)3−(−1)=124\frac{7 - (-5)}{3 - (-1)} = \frac{12}{4}; putting the xx-change on top gives the trap 13\frac{1}{3}.

    So the slope is 3.

  2. Q2Easy

    The point PP divides the line segment joining A(2,−3)A(2, -3) and B(8,9)B(8, 9) internally in the ratio 1:21 : 2. What is the distance of PP from the origin?

    Correct answer17\sqrt{17}

    Why

    Cut the segment into three equal steps: a 1:21 : 2 split is one step out from AA.

    One step is 13\frac{1}{3} of (8−2,9+3)=(6,12)(8 - 2, 9 + 3) = (6, 12), which is (2,4)(2, 4), so P=(2+2,−3+4)=(4,1)P = (2 + 2, -3 + 4) = (4, 1).

    Its distance from the origin is 42+12=17\sqrt{4^2 + 1^2} = \sqrt{17}.

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