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Samacheer Kalvi 10th Maths Exercise 5.2 Solutions

10th Maths Exercise 5.2 solutions (Chapter 5 Coordinate Geometry) with step-by-step answers for all 17 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance11 answers · full steps below
  1. Q1 (i)Slope is undefined (not defined)
  2. Q1 (ii)m=0m=0
  3. Q2 (i)θ=0∘\theta=0^\circ
  4. Q2 (ii)θ=45∘\theta=45^\circ
  5. Q3 (i)15\dfrac{1}{\sqrt5}
  6. Q3 (ii)−cot⁡θ-\cot\theta
  7. Q433
  8. Q6a=7a=7
  9. Q7a=172a=\dfrac{17}{2}
  10. Q8x=4x=4
  11. Q11x=5, y=2x=5,\ y=2

Q1. What is the slope of a line whose inclination with positive direction …

What is the slope of a line whose inclination with positive direction of xx-axis is

(i)

90∘90^\circ

  1. Slope m=tan⁡θm=\tan\theta.

  2. Here θ=90∘\theta=90^\circ, and tan⁡90∘\tan 90^\circ is not defined.

  3. So the line is vertical and its slope is undefined.

Answer

Slope is undefined (not defined)

(ii)

0∘0^\circ

  1. Slope m=tan⁡θm=\tan\theta.

  2. Here θ=0∘\theta=0^\circ, so m=tan⁡0∘=0m=\tan 0^\circ=0.

Answer

m=0m=0

Q2. What is the inclination of a line whose slope is

(i)

00

  1. Let θ\theta be the inclination.

  2. tan⁡θ=0\tan\theta=0

  3. So θ=0∘\theta=0^\circ.

Answer

θ=0∘\theta=0^\circ

(ii)

11

  1. Let θ\theta be the inclination.

  2. tan⁡θ=1\tan\theta=1

  3. So θ=45∘\theta=45^\circ.

Answer

θ=45∘\theta=45^\circ

Q3. Find the slope of a line joining the points

(i)

(5,5)\left(5,\sqrt{5}\right) with the origin

  1. The points are (5,5)(5,\sqrt5) and the origin (0,0)(0,0).

  2. Slope =y2−y1x2−x1=5−05−0=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\sqrt5-0}{5-0}

  3. =55=15=\dfrac{\sqrt5}{5}=\dfrac{1}{\sqrt5}.

Answer

15\dfrac{1}{\sqrt5}

(ii)

(sin⁡θ,−cos⁡θ)(\sin\theta, -\cos\theta) and (−sin⁡θ,cos⁡θ)(-\sin\theta, \cos\theta)

  1. Slope =y2−y1x2−x1=cos⁡θ−(−cos⁡θ)−sin⁡θ−sin⁡θ=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\cos\theta-(-\cos\theta)}{-\sin\theta-\sin\theta}

  2. =2cos⁡θ−2sin⁡θ=\dfrac{2\cos\theta}{-2\sin\theta}

  3. =−cot⁡θ=-\cot\theta.

Answer

−cot⁡θ-\cot\theta

Q4. What is the slope of a line perpendicular to the line joining …

What is the slope of a line perpendicular to the line joining A(5,1)A(5,1) and PP where PP is the mid-point of the segment joining (4,2)(4,2) and (−6,4)(-6,4).

  1. PP = mid-point of (4,2)(4,2) and (−6,4)(-6,4)

  2. P=(4−62,2+42)=(−1,3)P=\left(\dfrac{4-6}{2},\dfrac{2+4}{2}\right)=(-1,3)

  3. Slope of AP=3−1−1−5=2−6=−13AP=\dfrac{3-1}{-1-5}=\dfrac{2}{-6}=-\dfrac13

  4. If mm is the slope of the perpendicular line, m×(−13)=−1m\times\left(-\dfrac13\right)=-1

  5. m=3m=3.

Answer

33

Q5. Show that the given points are collinear: (−3,−4)(-3,-4) , (7,2)(7,2) and (12,5)(12,5)

  1. Let A(−3,−4)A(-3,-4), B(7,2)B(7,2), C(12,5)C(12,5).

  2. Slope of AB=2−(−4)7−(−3)=610=35AB=\dfrac{2-(-4)}{7-(-3)}=\dfrac{6}{10}=\dfrac35

  3. Slope of BC=5−212−7=35BC=\dfrac{5-2}{12-7}=\dfrac35

  4. Slope of ABAB = slope of BCBC, and BB is common to both.

  5. So AA, BB, CC lie on the same straight line.

Answer

Slope of ABAB = slope of BCBC = 35\dfrac35, hence the points are collinear.

Q6. If the three points (3,−1)(3,-1) , (a,3)(a,3) and (1,−3)(1,-3) are collinear, find …

If the three points (3,−1)(3,-1), (a,3)(a,3) and (1,−3)(1,-3) are collinear, find the value of aa.

Show steps (6)

  1. Let A(3,−1)A(3,-1), B(a,3)B(a,3), C(1,−3)C(1,-3).

  2. Slope of AC=−3−(−1)1−3=−2−2=1AC=\dfrac{-3-(-1)}{1-3}=\dfrac{-2}{-2}=1

  3. Slope of AB=3−(−1)a−3=4a−3AB=\dfrac{3-(-1)}{a-3}=\dfrac{4}{a-3}

  4. For collinear points, slope of ABAB = slope of ACAC

  5. 4a−3=1⇒a−3=4\dfrac{4}{a-3}=1\Rightarrow a-3=4

  6. a=7a=7.

Answer

a=7a=7

Q7. The line through the points (−2,a)(-2,a) and (9,3)(9,3) has slope −12-\dfrac{1}{2} . …

The line through the points (−2,a)(-2,a) and (9,3)(9,3) has slope −12-\dfrac{1}{2}. Find the value of aa.

  1. Slope =3−a9−(−2)=3−a11=\dfrac{3-a}{9-(-2)}=\dfrac{3-a}{11}

  2. Given slope =−12=-\dfrac12

  3. 3−a11=−12\dfrac{3-a}{11}=-\dfrac12

  4. 2(3−a)=−11⇒6−2a=−112(3-a)=-11\Rightarrow 6-2a=-11

  5. 2a=172a=17, so a=172a=\dfrac{17}{2}.

Answer

a=172a=\dfrac{17}{2}

Q8. The line through the points (−2,6)(-2,6) and (4,8)(4,8) is perpendicular to the …

The line through the points (−2,6)(-2,6) and (4,8)(4,8) is perpendicular to the line through the points (8,12)(8,12) and (x,24)(x,24). Find the value of xx.

Show steps (6)

  1. Slope of the first line m1=8−64−(−2)=26=13m_1=\dfrac{8-6}{4-(-2)}=\dfrac{2}{6}=\dfrac13

  2. Slope of the second line m2=24−12x−8=12x−8m_2=\dfrac{24-12}{x-8}=\dfrac{12}{x-8}

  3. Lines are perpendicular, so m1m2=−1m_1m_2=-1

  4. 13×12x−8=−1\dfrac13\times\dfrac{12}{x-8}=-1

  5. 4x−8=−1⇒x−8=−4\dfrac{4}{x-8}=-1\Rightarrow x-8=-4

  6. x=4x=4.

Answer

x=4x=4

Q9. Show that the given points form a right angled triangle.

(i)

A(1,−4)A(1,-4), B(2,−3)B(2,-3) and C(4,−7)C(4,-7)

  1. Slope of AB=−3+42−1=1AB=\dfrac{-3+4}{2-1}=1

  2. Slope of BC=−7+34−2=−42=−2BC=\dfrac{-7+3}{4-2}=\dfrac{-4}{2}=-2

  3. Slope of AC=−7+44−1=−33=−1AC=\dfrac{-7+4}{4-1}=\dfrac{-3}{3}=-1

  4. Slope of AB×AB\times slope of AC=(1)(−1)=−1AC=(1)(-1)=-1

  5. So AB⊥ACAB\perp AC, that is ∠A=90∘\angle A=90^\circ.

Answer

Slope AB×AB\times slope AC=−1AC=-1, so ∠A=90∘\angle A=90^\circ and △ABC\triangle ABC is right angled.

(ii)

L(0,5)L(0,5), M(9,12)M(9,12) and N(3,14)N(3,14)

  1. Slope of LM=12−59−0=79LM=\dfrac{12-5}{9-0}=\dfrac79

  2. Slope of MN=14−123−9=2−6=−13MN=\dfrac{14-12}{3-9}=\dfrac{2}{-6}=-\dfrac13

  3. Slope of LN=14−53−0=93=3LN=\dfrac{14-5}{3-0}=\dfrac93=3

  4. Slope of MN×MN\times slope of LN=(−13)(3)=−1LN=\left(-\dfrac13\right)(3)=-1

  5. So MN⊥LNMN\perp LN, that is ∠N=90∘\angle N=90^\circ.

Answer

Slope MN×MN\times slope LN=−1LN=-1, so ∠N=90∘\angle N=90^\circ and △LMN\triangle LMN is right angled.

Q10. Show that the given points form a parallelogram : A(2.5,3.5)A(2.5, 3.5) , B(10,−4)B(10,-4) …

Show that the given points form a parallelogram : A(2.5,3.5)A(2.5, 3.5), B(10,−4)B(10,-4), C(2.5,−2.5)C(2.5,-2.5) and D(−5,5)D(-5,5)

Show steps (7)

  1. Slope of AB=−4−3.510−2.5=−7.57.5=−1AB=\dfrac{-4-3.5}{10-2.5}=\dfrac{-7.5}{7.5}=-1

  2. Slope of CD=5−(−2.5)−5−2.5=7.5−7.5=−1CD=\dfrac{5-(-2.5)}{-5-2.5}=\dfrac{7.5}{-7.5}=-1

  3. Slope of BC=−2.5+42.5−10=1.5−7.5=−15BC=\dfrac{-2.5+4}{2.5-10}=\dfrac{1.5}{-7.5}=-\dfrac15

  4. Slope of AD=5−3.5−5−2.5=1.5−7.5=−15AD=\dfrac{5-3.5}{-5-2.5}=\dfrac{1.5}{-7.5}=-\dfrac15

  5. Slope of ABAB = slope of CDCD ⇒AB∥CD\Rightarrow AB\parallel CD

  6. Slope of BCBC = slope of ADAD ⇒BC∥AD\Rightarrow BC\parallel AD

  7. Both pairs of opposite sides are parallel.

Answer

Both pairs of opposite sides have equal slopes, hence ABCDABCD is a parallelogram.

Q11. If the points A(2,2)A(2,2) , B(−2,−3)B(-2,-3) , C(1,−3)C(1,-3) and D(x,y)D(x, y) form a …

If the points A(2,2)A(2,2), B(−2,−3)B(-2,-3), C(1,−3)C(1,-3) and D(x,y)D(x, y) form a parallelogram then find the value of xx and yy.

Show steps (6)

  1. In parallelogram ABCDABCD, AD∥BCAD\parallel BC and AB∥DCAB\parallel DC.

  2. Slope of BC=−3−(−3)1−(−2)=03=0BC=\dfrac{-3-(-3)}{1-(-2)}=\dfrac{0}{3}=0

  3. Slope of AD=y−2x−2AD=\dfrac{y-2}{x-2}; equate: y−2x−2=0⇒y=2\dfrac{y-2}{x-2}=0\Rightarrow y=2

  4. Slope of AB=−3−2−2−2=−5−4=54AB=\dfrac{-3-2}{-2-2}=\dfrac{-5}{-4}=\dfrac54

  5. Slope of DC=−3−y1−x=−51−xDC=\dfrac{-3-y}{1-x}=\dfrac{-5}{1-x}

  6. −51−x=54⇒1−x=−4⇒x=5\dfrac{-5}{1-x}=\dfrac54\Rightarrow 1-x=-4\Rightarrow x=5.

Answer

x=5, y=2x=5,\ y=2

Q12. Let A(3,−4)A(3,-4) , B(9,−4)B(9,-4) , C(5,−7)C(5,-7) and D(7,−7)D(7,-7) . Show that ABCDABCD …

Let A(3,−4)A(3,-4), B(9,−4)B(9,-4), C(5,−7)C(5,-7) and D(7,−7)D(7,-7). Show that ABCDABCD is a trapezium.

Show steps (7)

  1. Slope of AB=−4−(−4)9−3=06=0AB=\dfrac{-4-(-4)}{9-3}=\dfrac{0}{6}=0

  2. Slope of CD=−7−(−7)7−5=02=0CD=\dfrac{-7-(-7)}{7-5}=\dfrac{0}{2}=0

  3. So AB∥CDAB\parallel CD.

  4. Slope of BC=−7+45−9=−3−4=34BC=\dfrac{-7+4}{5-9}=\dfrac{-3}{-4}=\dfrac34

  5. Slope of AD=−7+47−3=−34AD=\dfrac{-7+4}{7-3}=\dfrac{-3}{4}

  6. Slopes of BCBC and ADAD are not equal, so BCBC and ADAD are not parallel.

  7. Exactly one pair of opposite sides is parallel.

Answer

AB∥CDAB\parallel CD but BC∦ADBC\nparallel AD, hence ABCDABCD is a trapezium.

Q13. A quadrilateral has vertices at A(−4,−2)A(-4,-2) , B(5,−1)B(5,-1) , C(6,5)C(6,5) and D(−7,6)D(-7,6) …

A quadrilateral has vertices at A(−4,−2)A(-4,-2), B(5,−1)B(5,-1), C(6,5)C(6,5) and D(−7,6)D(-7,6). Show that the mid-points of its sides form a parallelogram.

Show steps (10)

  1. Let P,Q,R,SP,Q,R,S be the mid-points of AB,BC,CD,DAAB,BC,CD,DA.

  2. P=(−4+52,−2−12)=(12,−32)P=\left(\dfrac{-4+5}{2},\dfrac{-2-1}{2}\right)=\left(\dfrac12,-\dfrac32\right)

  3. Q=(5+62,−1+52)=(112,2)Q=\left(\dfrac{5+6}{2},\dfrac{-1+5}{2}\right)=\left(\dfrac{11}{2},2\right)

  4. R=(6−72,5+62)=(−12,112)R=\left(\dfrac{6-7}{2},\dfrac{5+6}{2}\right)=\left(-\dfrac12,\dfrac{11}{2}\right)

  5. S=(−7−42,6−22)=(−112,2)S=\left(\dfrac{-7-4}{2},\dfrac{6-2}{2}\right)=\left(-\dfrac{11}{2},2\right)

  6. Slope of PQ=2+32112−12=7/25=710PQ=\dfrac{2+\frac32}{\frac{11}{2}-\frac12}=\dfrac{7/2}{5}=\dfrac{7}{10}

  7. Slope of SR=112−2−12+112=7/25=710SR=\dfrac{\frac{11}{2}-2}{-\frac12+\frac{11}{2}}=\dfrac{7/2}{5}=\dfrac{7}{10}

  8. Slope of QR=112−2−12−112=7/2−6=−712QR=\dfrac{\frac{11}{2}-2}{-\frac12-\frac{11}{2}}=\dfrac{7/2}{-6}=-\dfrac{7}{12}

  9. Slope of PS=2+32−112−12=7/2−6=−712PS=\dfrac{2+\frac32}{-\frac{11}{2}-\frac12}=\dfrac{7/2}{-6}=-\dfrac{7}{12}

  10. PQ∥SRPQ\parallel SR and QR∥PSQR\parallel PS.

Answer

Both pairs of opposite sides of PQRSPQRS have equal slopes, hence the mid-points form a parallelogram.

Quick check

  1. A straight line has equation 8y=4x+218y = 4x + 21. Which of the following is true

    1. (1) The slope is 0.5 and the yy intercept is 2.6
    2. (2) The slope is 5 and the yy intercept is 1.6
    3. (3) The slope is 0.5 and the yy intercept is 1.6
    4. (4) The slope is 5 and the yy intercept is 2.6
    Show answer

    Answer: (1) The slope is 0.5 and the yy intercept is 2.6

    1. 8y=4x+21⇒y=48x+2188y=4x+21\Rightarrow y=\dfrac{4}{8}x+\dfrac{21}{8}.

    2. y=0.5x+2.625y=0.5x+2.625.

    3. Slope =0.5=0.5, yy intercept =2.625≈2.6=2.625\approx 2.6.

    Question 1 of 5

  2. If (5,7)(5,7), (3,p)(3,p) and (6,6)(6,6) are collinear, then the value of pp is

    1. (1) 33
    2. (2) 66
    3. (3) 99
    4. (4) 1212
    Show answer

    Answer: (3) 99

    1. Slope of the line joining (5,7)(5,7) and (6,6)(6,6) is 6−76−5=−1\dfrac{6-7}{6-5}=-1.

    2. For collinearity, slope of (5,7)(5,7) and (3,p)(3,p) is also −1-1.

    3. p−73−5=−1⇒p−7=2\dfrac{p-7}{3-5}=-1\Rightarrow p-7=2

    4. p=9p=9

    Question 2 of 5

  3. If AA is a point on the YY axis whose ordinate is 8 and BB is a point on the XX axis whose abscissae is 5 then the equation of the line ABAB is

    1. (1) 8x+5y=408x + 5y = 40
    2. (2) 8x−5y=408x - 5y = 40
    3. (3) x=8x = 8
    4. (4) y=5y = 5
    Show answer

    Answer: (1) 8x+5y=408x + 5y = 40

    1. AA is on the Y axis with ordinate 88, so A(0,8)A(0,8); BB is on the X axis with abscissa 55, so B(5,0)B(5,0).

    2. Intercept form: x5+y8=1\dfrac{x}{5}+\dfrac{y}{8}=1.

    3. Multiplying by 4040: 8x+5y=408x+5y=40.

    Question 3 of 5

  4. The equation of a line passing through the origin and perpendicular to the line 7x−3y+4=07x - 3y + 4 = 0 is

    1. (1) 7x−3y+4=07x - 3y + 4 = 0
    2. (2) 3x−7y+4=03x - 7y + 4 = 0
    3. (3) 3x+7y=03x + 7y = 0
    4. (4) 7x−3y=07x - 3y = 0
    Show answer

    Answer: (3) 3x+7y=03x + 7y = 0

    1. Slope of 7x−3y+4=07x-3y+4=0 is −7−3=73\dfrac{-7}{-3}=\dfrac{7}{3}.

    2. Slope of the perpendicular line =−37=-\dfrac{3}{7}.

    3. Through the origin: y=−37x⇒3x+7y=0y=-\dfrac{3}{7}x\Rightarrow 3x+7y=0.

    Question 4 of 5

  5. A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the YY axis. The path travelled by the man is

    1. (1) x=10x = 10
    2. (2) y=10y = 10
    3. (3) x=0x = 0
    4. (4) y=0y = 0
    Show answer

    Answer: (1) x=10x = 10

    1. The wall is the Y axis, and his distance from it is always 1010 units.

    2. So his xx coordinate is always 1010.

    3. Path is the line x=10x=10.

    Question 5 of 5
Chapter 5 formulas
  • Area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃)

    Area=12{(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)} sq.units\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right\}\ \text{sq.units}

  • Collinearity condition (area)

    Area of △ABC=0orx1y2+x2y3+x3y1=x2y1+x3y2+x1y3\text{Area of }\triangle ABC=0\quad\text{or}\quad x_1y_2+x_2y_3+x_3y_1=x_2y_1+x_3y_2+x_1y_3

  • Collinearity condition (slope)

    slope of AB=slope of BC=slope of AC\text{slope of }AB=\text{slope of }BC=\text{slope of }AC

  • Area of a quadrilateral with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄)

    Area=12{(x1y2+x2y3+x3y4+x4y1)−(x2y1+x3y2+x4y3+x1y4)} sq.units\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\right\}\ \text{sq.units}

  • Slope of a line making angle θ with the positive direction of the X axis

    m=tan⁡θm=\tan\theta

  • Slope of a line joining two distinct points A(x₁,y₁) and B(x₂,y₂)

    m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}

  • Slope of the line ax + by + c = 0

    m=−abm=\frac{-a}{b}

  • General form of a straight line

    ax+by+c=0ax+by+c=0

  • Point-slope form

    y−y1=m(x−x1)y-y_1=m(x-x_1)

  • Slope-intercept form

    y=mx+cy=mx+c

  • Two point form

    y−y1y2−y1=x−x1x2−x1\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}

  • Intercept form

    xa+yb=1\frac{x}{a}+\frac{y}{b}=1

  • Line parallel to the Y axis

    x=cx=c

  • Line parallel to the X axis

    y=by=b

  • Condition for two straight lines to be parallel

    m1=m2m_1=m_2

  • Condition for two straight lines to be perpendicular

    m1×m2=−1m_1\times m_2=-1

Answers checked against the Samacheer Kalvi textbook · Updated

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