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Samacheer Kalvi 10th Maths Exercise 5.4 Solutions

10th Maths Exercise 5.4 solutions (Chapter 5 Coordinate Geometry) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1 (i)Slope m=0m = 0
  2. Q1 (ii)Slope is undefined (line parallel to Y axis)
  3. Q2 (i)Slope =0.7=710= 0.7 = \dfrac{7}{10}
  4. Q2 (ii)Slope =0= 0
  5. Q3 (i)The lines are parallel (m1=m2=−43m_1=m_2=-\dfrac{4}{3})
  6. Q3 (ii)The lines are perpendicular (m1m2=−1m_1m_2=-1)
  7. Q4p=4p = 4
  8. Q53x+4y+7=03x+4y+7=0
  9. Q62x+5y−2=02x+5y-2=0
  10. Q7Altitude through AA: 2x+5y+6=02x+5y+6=0; Altitude through BB: 5x+y−48=05x+y-48=0
  11. Q85x−3y−8=05x-3y-8=0
  12. Q913x+5y−18=013x+5y-18=0
  13. Q1049x+28y−156=049x+28y-156=0
  14. Q1131x+15y+30=031x+15y+30=0
  15. Q124x+13y−9=04x+13y-9=0

Q1. Find the slope of the following straight lines

(i)

5y−3=05y - 3 = 0

  1. 5y−3=05y-3=0 has coefficient of x=0x = 0, coefficient of y=5y = 5.

  2. m=−coefficient of xcoefficient of y=−05=0m=\dfrac{-\text{coefficient of }x}{\text{coefficient of }y}=\dfrac{-0}{5}=0.

  3. The line is parallel to the X axis.

Answer

Slope m=0m = 0

(ii)

7x−317=07x - \dfrac{3}{17} = 0

  1. 7x−317=07x-\dfrac{3}{17}=0 has coefficient of x=7x = 7, coefficient of y=0y = 0.

  2. m=−coefficient of xcoefficient of y=−70m=\dfrac{-\text{coefficient of }x}{\text{coefficient of }y}=\dfrac{-7}{0}, which is not defined.

  3. The line x=3119x=\dfrac{3}{119} is parallel to the Y axis, so its slope is undefined.

Answer

Slope is undefined (line parallel to Y axis)

Q2. Find the slope of the line which is

(i)

parallel to y=0.7x−11y = 0.7x - 11

  1. y=0.7x−11y=0.7x-11 is of the form y=mx+cy=mx+c.

  2. So its slope is 0.70.7.

  3. Parallel lines have equal slopes.

Answer

Slope =0.7=710= 0.7 = \dfrac{7}{10}

(ii)

perpendicular to the line x=−11x = -11

  1. x=−11x=-11 is a line parallel to the Y axis.

  2. A line perpendicular to it is parallel to the X axis.

  3. Every line parallel to the X axis has slope 00.

Answer

Slope =0= 0

Q3. Check whether the given lines are parellel or perpendicular

(i)

x3+y4+17=0\dfrac{x}{3} + \dfrac{y}{4} + \dfrac{1}{7} = 0 and 2x3+y2+110=0\dfrac{2x}{3} + \dfrac{y}{2} + \dfrac{1}{10} = 0

  1. For x3+y4+17=0\dfrac{x}{3}+\dfrac{y}{4}+\dfrac{1}{7}=0, m1=−1/31/4=−43m_1=\dfrac{-1/3}{1/4}=-\dfrac{4}{3}.

  2. For 2x3+y2+110=0\dfrac{2x}{3}+\dfrac{y}{2}+\dfrac{1}{10}=0, m2=−2/31/2=−43m_2=\dfrac{-2/3}{1/2}=-\dfrac{4}{3}.

  3. m1=m2m_1=m_2, so the lines are parallel.

Answer

The lines are parallel (m1=m2=−43m_1=m_2=-\dfrac{4}{3})

(ii)

5x+23y+14=05x + 23y + 14 = 0 and 23x−5y+9=023x - 5y + 9 = 0

  1. For 5x+23y+14=05x+23y+14=0, m1=−523m_1=-\dfrac{5}{23}.

  2. For 23x−5y+9=023x-5y+9=0, m2=−23−5=235m_2=-\dfrac{23}{-5}=\dfrac{23}{5}.

  3. m1×m2=−523×235=−1m_1\times m_2=-\dfrac{5}{23}\times\dfrac{23}{5}=-1.

Answer

The lines are perpendicular (m1m2=−1m_1m_2=-1)

Q4. If the straight lines 12y=−(p+3)x+1212y = -(p+3)x + 12 , 12x−7y=1612x - 7y = 16 are perpendicular then find 'pp'.

Show steps (6)

  1. 12y=−(p+3)x+12⇒(p+3)x+12y−12=012y=-(p+3)x+12 \Rightarrow (p+3)x+12y-12=0, so m1=−p+312m_1=-\dfrac{p+3}{12}.

  2. 12x−7y=16⇒12x−7y−16=012x-7y=16 \Rightarrow 12x-7y-16=0, so m2=−12−7=127m_2=-\dfrac{12}{-7}=\dfrac{12}{7}.

  3. Perpendicular ⇒m1m2=−1\Rightarrow m_1m_2=-1:

  4. −p+312×127=−1-\dfrac{p+3}{12}\times\dfrac{12}{7}=-1

  5. −p+37=−1⇒p+3=7-\dfrac{p+3}{7}=-1 \Rightarrow p+3=7

  6. p=4p=4

Answer

p=4p = 4

Q5. Find the equation of a straight line passing through the point P(−5,2)P(-5,2) …

Find the equation of a straight line passing through the point P(−5,2)P(-5,2) and parallel to the line joining the points Q(3,−2)Q(3,-2) and R(−5,4)R(-5,4).

Show steps (6)

  1. Slope of QRQR with Q(3,−2)Q(3,-2), R(−5,4)R(-5,4):

  2. m=4−(−2)−5−3=6−8=−34m=\dfrac{4-(-2)}{-5-3}=\dfrac{6}{-8}=-\dfrac{3}{4}.

  3. Required line is parallel, so its slope is −34-\dfrac{3}{4} and it passes through P(−5,2)P(-5,2).

  4. y−2=−34(x+5)y-2=-\dfrac{3}{4}(x+5)

  5. 4y−8=−3x−154y-8=-3x-15

  6. 3x+4y+7=03x+4y+7=0

Answer

3x+4y+7=03x+4y+7=0

Q6. Find the equation of a line passing through (6,−2)(6,-2) and perpendicular to …

Find the equation of a line passing through (6,−2)(6,-2) and perpendicular to the line joining the points (6,7)(6,7) and (2,−3)(2,-3).

Show steps (7)

  1. Slope of the line joining (6,7)(6,7) and (2,−3)(2,-3):

  2. m=−3−72−6=−10−4=52m=\dfrac{-3-7}{2-6}=\dfrac{-10}{-4}=\dfrac{5}{2}.

  3. Slope of the perpendicular line =−25=-\dfrac{2}{5}.

  4. It passes through (6,−2)(6,-2):

  5. y+2=−25(x−6)y+2=-\dfrac{2}{5}(x-6)

  6. 5y+10=−2x+125y+10=-2x+12

  7. 2x+5y−2=02x+5y-2=0

Answer

2x+5y−2=02x+5y-2=0

Q7. A(−3,0)A(-3,0) B(10,−2)B(10,-2) and C(12,3)C(12,3) are the vertices of ΔABC\Delta ABC . Find the …

A(−3,0)A(-3,0) B(10,−2)B(10,-2) and C(12,3)C(12,3) are the vertices of ΔABC\Delta ABC. Find the equation of the altitude through AA and BB.

Show steps (10)

  1. Altitude through AA is perpendicular to BCBC.

  2. Slope of BCBC, B(10,−2)B(10,-2), C(12,3)C(12,3): 3−(−2)12−10=52\dfrac{3-(-2)}{12-10}=\dfrac{5}{2}.

  3. Slope of altitude =−25=-\dfrac{2}{5}, through A(−3,0)A(-3,0):

  4. y−0=−25(x+3)⇒5y=−2x−6y-0=-\dfrac{2}{5}(x+3)\Rightarrow 5y=-2x-6

  5. 2x+5y+6=02x+5y+6=0

  6. Altitude through BB is perpendicular to ACAC.

  7. Slope of ACAC, A(−3,0)A(-3,0), C(12,3)C(12,3): 3−012+3=15\dfrac{3-0}{12+3}=\dfrac{1}{5}.

  8. Slope of altitude =−5=-5, through B(10,−2)B(10,-2):

  9. y+2=−5(x−10)⇒y+2=−5x+50y+2=-5(x-10)\Rightarrow y+2=-5x+50

  10. 5x+y−48=05x+y-48=0

Answer

Altitude through AA: 2x+5y+6=02x+5y+6=0; Altitude through BB: 5x+y−48=05x+y-48=0

Q8. Find the equation of the perpendicular bisector of the line joining the …

Find the equation of the perpendicular bisector of the line joining the points A(−4,2)A(-4,2) and B(6,−4)B(6,-4).

Show steps (7)

  1. Midpoint of A(−4,2)A(-4,2), B(6,−4)B(6,-4):

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

  3. Slope of AB=−4−26+4=−610=−35AB=\dfrac{-4-2}{6+4}=\dfrac{-6}{10}=-\dfrac{3}{5}.

  4. Slope of perpendicular bisector =53=\dfrac{5}{3}, through (1,−1)(1,-1):

  5. y+1=53(x−1)y+1=\dfrac{5}{3}(x-1)

  6. 3y+3=5x−53y+3=5x-5

  7. 5x−3y−8=05x-3y-8=0

Answer

5x−3y−8=05x-3y-8=0

Q9. Find the equation of a straight line through the intersection of lines …

Find the equation of a straight line through the intersection of lines 7x+3y=107x + 3y = 10, 5x−4y=15x - 4y = 1 and parallel to the line 13x+5y+12=013x + 5y + 12 = 0

Show steps (8)

  1. Solve 7x+3y=107x+3y=10 and 5x−4y=15x-4y=1.

  2. (7x+3y=10)×4: 28x+12y=40(7x+3y=10)\times 4:\ 28x+12y=40

  3. (5x−4y=1)×3: 15x−12y=3(5x-4y=1)\times 3:\ 15x-12y=3

  4. Adding, 43x=43⇒x=143x=43\Rightarrow x=1, then 7+3y=10⇒y=17+3y=10\Rightarrow y=1.

  5. Point of intersection (1,1)(1,1).

  6. Line parallel to 13x+5y+12=013x+5y+12=0 is 13x+5y+k=013x+5y+k=0.

  7. 13(1)+5(1)+k=0⇒k=−1813(1)+5(1)+k=0\Rightarrow k=-18

  8. 13x+5y−18=013x+5y-18=0

Answer

13x+5y−18=013x+5y-18=0

Q10. Find the equation of a straight line through the intersection of lines …

Find the equation of a straight line through the intersection of lines 5x−6y=25x - 6y = 2, 3x+2y=103x + 2y = 10 and perpendicular to the line 4x−7y+13=04x - 7y + 13 = 0

Show steps (9)

  1. Solve 5x−6y=25x-6y=2 and 3x+2y=103x+2y=10.

  2. (3x+2y=10)×3: 9x+6y=30(3x+2y=10)\times 3:\ 9x+6y=30

  3. Adding to 5x−6y=25x-6y=2: 14x=32⇒x=16714x=32\Rightarrow x=\dfrac{16}{7}.

  4. Then 2y=10−487=227⇒y=1172y=10-\dfrac{48}{7}=\dfrac{22}{7}\Rightarrow y=\dfrac{11}{7}.

  5. Point of intersection (167,117)\left(\dfrac{16}{7},\dfrac{11}{7}\right).

  6. Line perpendicular to 4x−7y+13=04x-7y+13=0 is 7x+4y+k=07x+4y+k=0.

  7. 7(167)+4(117)+k=07\left(\dfrac{16}{7}\right)+4\left(\dfrac{11}{7}\right)+k=0

  8. 16+447+k=0⇒k=−156716+\dfrac{44}{7}+k=0\Rightarrow k=-\dfrac{156}{7}

  9. 7x+4y−1567=0⇒49x+28y−156=07x+4y-\dfrac{156}{7}=0\Rightarrow 49x+28y-156=0

Answer

49x+28y−156=049x+28y-156=0

Q11. Find the equation of a straight line joining the point of intersection …

Find the equation of a straight line joining the point of intersection of 3x+y+2=03x + y + 2 = 0 and x−2y−4=0x - 2y - 4 = 0 to the point of intersection of 7x−3y=−127x - 3y = -12 and 2y=x+32y = x + 3

Show steps (10)

  1. Solve 3x+y+2=03x+y+2=0 and x−2y−4=0x-2y-4=0.

  2. From the first, y=−3x−2y=-3x-2; substituting: x−2(−3x−2)−4=0x-2(-3x-2)-4=0

  3. 7x=0⇒x=0, y=−27x=0\Rightarrow x=0,\ y=-2. First point A(0,−2)A(0,-2).

  4. Solve 7x−3y=−127x-3y=-12 and x=2y−3x=2y-3.

  5. 7(2y−3)−3y=−12⇒11y=9⇒y=9117(2y-3)-3y=-12\Rightarrow 11y=9\Rightarrow y=\dfrac{9}{11}

  6. x=2(911)−3=−1511x=2\left(\dfrac{9}{11}\right)-3=-\dfrac{15}{11}. Second point B(−1511,911)B\left(-\dfrac{15}{11},\dfrac{9}{11}\right).

  7. Slope =911+2−1511−0=31/11−15/11=−3115=\dfrac{\frac{9}{11}+2}{-\frac{15}{11}-0}=\dfrac{31/11}{-15/11}=-\dfrac{31}{15}.

  8. Through (0,−2)(0,-2): y+2=−3115xy+2=-\dfrac{31}{15}x

  9. 15y+30=−31x15y+30=-31x

  10. 31x+15y+30=031x+15y+30=0

Answer

31x+15y+30=031x+15y+30=0

Q12. Find the equation of a straight line through the point of intersection …

Find the equation of a straight line through the point of intersection of the lines 8x+3y=188x + 3y = 18, 4x+5y=94x + 5y = 9 and bisecting the line segment joining the points (5,−4)(5,-4) and (−7,6)(-7,6).

Show steps (9)

  1. Solve 8x+3y=188x+3y=18 and 4x+5y=94x+5y=9.

  2. (4x+5y=9)×2: 8x+10y=18(4x+5y=9)\times 2:\ 8x+10y=18

  3. Subtracting: 7y=0⇒y=07y=0\Rightarrow y=0, then 8x=18⇒x=948x=18\Rightarrow x=\dfrac{9}{4}.

  4. Point of intersection (94,0)\left(\dfrac{9}{4},0\right).

  5. Midpoint of (5,−4)(5,-4) and (−7,6)(-7,6) is (5−72,−4+62)=(−1,1)\left(\dfrac{5-7}{2},\dfrac{-4+6}{2}\right)=(-1,1).

  6. Slope =1−0−1−94=1−134=−413=\dfrac{1-0}{-1-\frac{9}{4}}=\dfrac{1}{-\frac{13}{4}}=-\dfrac{4}{13}.

  7. Through (94,0)\left(\dfrac{9}{4},0\right): y=−413(x−94)y=-\dfrac{4}{13}\left(x-\dfrac{9}{4}\right)

  8. 13y=−4x+913y=-4x+9

  9. 4x+13y−9=04x+13y-9=0

Answer

4x+13y−9=04x+13y-9=0

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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