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Samacheer Kalvi 10th Maths Exercise 5.3 Solutions

10th Maths Exercise 5.3 solutions (Chapter 5 Coordinate Geometry) with step-by-step answers for all 22 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance22 answers · full steps below
  1. Q1 (i)2y+3=02y+3=0
  2. Q1 (ii)2x−5=02x-5=0
  3. Q2Slope =1=1, inclination =45∘=45^\circ, yy intercept =52=\dfrac{5}{2}
  4. Q3x−3 y−33=0x-\sqrt3\,y-3\sqrt3=0
  5. Q4Slope =3+32=\dfrac{3+\sqrt3}{2}, yy intercept =−3(1+3)2=\dfrac{-3(1+\sqrt3)}{2}
  6. Q5a=−5a=-5
  7. Q6x−y−16=0x-y-16=0
  8. Q7 (i)16x−15y−22=016x-15y-22=0
  9. Q7 (ii)4x−9y+19=04x-9y+19=0
  10. Q815x−11y+46=015x-11y+46=0
  11. Q9 (i)x+4y−14=0x+4y-14=0
  12. Q9 (ii)3x+5y−28=03x+5y-28=0
  13. Q105x+4y−3=05x+4y-3=0
  14. Q11 (i)11 MB
  15. Q11 (ii)7.57.5 seconds
  16. Q11 (iii)1010 seconds
  17. Q12 (i)3x−2y−12=03x-2y-12=0
  18. Q12 (ii)3x−20y+15=03x-20y+15=0
  19. Q13 (i)xx intercept =2=2, yy intercept =−3=-3
  20. Q13 (ii)xx intercept =−3=-3, yy intercept =−4=-4
  21. Q14 (i)5x+2y+3=05x+2y+3=0
  22. Q14 (ii)x+y+4=0x+y+4=0

Q1. Find the equation of a straight line passing through the mid-point of …

Find the equation of a straight line passing through the mid-point of a line segment joining the points (1,−5)(1,-5), (4,2)(4,2) and parallel to

(i)

XX axis

  1. Mid-point =(1+42,−5+22)=(52,−32)=\left(\dfrac{1+4}{2},\dfrac{-5+2}{2}\right)=\left(\dfrac{5}{2},\dfrac{-3}{2}\right)

  2. A line parallel to the XX axis has the form y=ky=k.

  3. It passes through the mid-point, so y=−32y=\dfrac{-3}{2}.

  4. 2y+3=02y+3=0

Answer

2y+3=02y+3=0

(ii)

YY axis

  1. Mid-point =(1+42,−5+22)=(52,−32)=\left(\dfrac{1+4}{2},\dfrac{-5+2}{2}\right)=\left(\dfrac{5}{2},\dfrac{-3}{2}\right)

  2. A line parallel to the YY axis has the form x=kx=k.

  3. It passes through the mid-point, so x=52x=\dfrac{5}{2}.

  4. 2x−5=02x-5=0

Answer

2x−5=02x-5=0

Q2. The equation of a straight line is 2(x−y)+5=02(x-y)+5=0 . Find its slope …

The equation of a straight line is 2(x−y)+5=02(x-y)+5=0. Find its slope, inclination and intercept on the YY axis.

Show steps (6)

  1. 2(x−y)+5=0⇒2x−2y+5=02(x-y)+5=0\Rightarrow 2x-2y+5=0

  2. Write in the form y=mx+cy=mx+c:

  3. 2y=2x+52y=2x+5

  4. y=x+52y=x+\dfrac{5}{2}

  5. Slope m=1m=1, yy intercept c=52c=\dfrac{5}{2}.

  6. tan⁡θ=1⇒θ=45∘\tan\theta=1\Rightarrow \theta=45^\circ

Answer

Slope =1=1, inclination =45∘=45^\circ, yy intercept =52=\dfrac{5}{2}

Q3. Find the equation of a line whose inclination is 30∘30^\circ and making …

Find the equation of a line whose inclination is 30∘30^\circ and making an intercept −3-3 on the YY axis.

  1. Slope m=tan⁡30∘=13m=\tan 30^\circ=\dfrac{1}{\sqrt3}, yy intercept c=−3c=-3.

  2. Slope-intercept form y=mx+cy=mx+c:

  3. y=13x−3y=\dfrac{1}{\sqrt3}x-3

  4. 3 y=x−33\sqrt3\,y=x-3\sqrt3

  5. x−3 y−33=0x-\sqrt3\,y-3\sqrt3=0

Answer

x−3 y−33=0x-\sqrt3\,y-3\sqrt3=0

Q4. Find the slope and yy intercept of 3x+(1−3)y=3\sqrt{3}x+(1-\sqrt{3})y=3 .

  1. 3 x+(1−3)y=3\sqrt3\,x+(1-\sqrt3)y=3

  2. (1−3)y=−3 x+3(1-\sqrt3)y=-\sqrt3\,x+3

  3. y=−31−3x+31−3y=\dfrac{-\sqrt3}{1-\sqrt3}x+\dfrac{3}{1-\sqrt3}

  4. Slope =−31−3×1+31+3=−3−3−2=3+32=\dfrac{-\sqrt3}{1-\sqrt3}\times\dfrac{1+\sqrt3}{1+\sqrt3}=\dfrac{-\sqrt3-3}{-2}=\dfrac{3+\sqrt3}{2}

  5. yy intercept =31−3×1+31+3=3(1+3)−2=−3(1+3)2=\dfrac{3}{1-\sqrt3}\times\dfrac{1+\sqrt3}{1+\sqrt3}=\dfrac{3(1+\sqrt3)}{-2}=\dfrac{-3(1+\sqrt3)}{2}

Answer

Slope =3+32=\dfrac{3+\sqrt3}{2}, yy intercept =−3(1+3)2=\dfrac{-3(1+\sqrt3)}{2}

Q5. Find the value of 'aa', if the line through (−2,3)(-2,3) and (8,5)(8,5) …

Find the value of 'aa', if the line through (−2,3)(-2,3) and (8,5)(8,5) is perpendicular to y=ax+2y=ax+2

Show steps (6)

  1. Slope of the line through (−2,3)(-2,3) and (8,5)(8,5):

  2. m1=5−38−(−2)=210=15m_1=\dfrac{5-3}{8-(-2)}=\dfrac{2}{10}=\dfrac{1}{5}

  3. Slope of y=ax+2y=ax+2 is m2=am_2=a.

  4. Perpendicular ⇒m1m2=−1\Rightarrow m_1m_2=-1

  5. 15×a=−1\dfrac{1}{5}\times a=-1

  6. a=−5a=-5

Answer

a=−5a=-5

Q6. The hill in the form of a right triangle has its foot …

The hill in the form of a right triangle has its foot at (19,3)(19,3). The inclination of the hill to the ground is 45∘45^\circ. Find the equation of the hill joining the foot and top.

Show steps (6)

  1. Inclination =45∘=45^\circ, so slope m=tan⁡45∘=1m=\tan45^\circ=1.

  2. Point on the line: (19,3)(19,3).

  3. Point-slope form y−y1=m(x−x1)y-y_1=m(x-x_1):

  4. y−3=1(x−19)y-3=1(x-19)

  5. y−3=x−19y-3=x-19

  6. x−y−16=0x-y-16=0

Answer

x−y−16=0x-y-16=0

Q7. Find the equation of a line through the given pair of points

(i)

(2,23)\left(2,\dfrac{2}{3}\right) and (−12,−2)\left(\dfrac{-1}{2},-2\right)

Show steps (7)

  1. Two-point form y−y1y2−y1=x−x1x2−x1\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1} with (2,23)\left(2,\dfrac23\right), (−12,−2)\left(\dfrac{-1}{2},-2\right):

  2. y−23−2−23=x−2−12−2\dfrac{y-\frac23}{-2-\frac23}=\dfrac{x-2}{-\frac12-2}

  3. y−23−83=x−2−52\dfrac{y-\frac23}{-\frac83}=\dfrac{x-2}{-\frac52}

  4. −52(y−23)=−83(x−2)-\dfrac52\left(y-\dfrac23\right)=-\dfrac83(x-2)

  5. Multiply by −6-6: 15(y−23)=16(x−2)15\left(y-\dfrac23\right)=16(x-2)

  6. 15y−10=16x−3215y-10=16x-32

  7. 16x−15y−22=016x-15y-22=0

Answer

16x−15y−22=016x-15y-22=0

(ii)

(2,3)(2,3) and (−7,−1)(-7,-1)

Show steps (6)

  1. Two-point form with (2,3)(2,3) and (−7,−1)(-7,-1):

  2. y−3−1−3=x−2−7−2\dfrac{y-3}{-1-3}=\dfrac{x-2}{-7-2}

  3. y−3−4=x−2−9\dfrac{y-3}{-4}=\dfrac{x-2}{-9}

  4. −9(y−3)=−4(x−2)-9(y-3)=-4(x-2)

  5. 9y−27=4x−89y-27=4x-8

  6. 4x−9y+19=04x-9y+19=0

Answer

4x−9y+19=04x-9y+19=0

Q8. A cat is located at the point (−6,−4)(-6,-4) in xyxy plane. A …

A cat is located at the point (−6,−4)(-6,-4) in xyxy plane. A bottle of milk is kept at (5,11)(5,11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.

Show steps (6)

  1. Shortest path is the straight line joining (−6,−4)(-6,-4) and (5,11)(5,11).

  2. y−(−4)11−(−4)=x−(−6)5−(−6)\dfrac{y-(-4)}{11-(-4)}=\dfrac{x-(-6)}{5-(-6)}

  3. y+415=x+611\dfrac{y+4}{15}=\dfrac{x+6}{11}

  4. 11(y+4)=15(x+6)11(y+4)=15(x+6)

  5. 11y+44=15x+9011y+44=15x+90

  6. 15x−11y+46=015x-11y+46=0

Answer

15x−11y+46=015x-11y+46=0

Q9. If the vertices of a ΔABC\Delta ABC are A(6,2)A(6,2) , B(−5,−1)B(-5,-1) and C(1,9)C(1,9) …

If the vertices of a ΔABC\Delta ABC are A(6,2)A(6,2), B(−5,−1)B(-5,-1) and C(1,9)C(1,9); through the vertex AA

(i)

find the equation of median

Show steps (8)

  1. Median from AA goes to the mid-point DD of BCBC.

  2. D=(−5+12,−1+92)=(−2,4)D=\left(\dfrac{-5+1}{2},\dfrac{-1+9}{2}\right)=(-2,4)

  3. Line through A(6,2)A(6,2) and D(−2,4)D(-2,4):

  4. y−24−2=x−6−2−6\dfrac{y-2}{4-2}=\dfrac{x-6}{-2-6}

  5. y−22=x−6−8\dfrac{y-2}{2}=\dfrac{x-6}{-8}

  6. −8(y−2)=2(x−6)-8(y-2)=2(x-6)

  7. −8y+16=2x−12-8y+16=2x-12

  8. x+4y−14=0x+4y-14=0

Answer

x+4y−14=0x+4y-14=0

(ii)

find the equation of altitude

  1. Slope of BC=9−(−1)1−(−5)=106=53BC=\dfrac{9-(-1)}{1-(-5)}=\dfrac{10}{6}=\dfrac53

  2. Altitude from AA is perpendicular to BCBC, so its slope =−35=\dfrac{-3}{5}.

  3. Through A(6,2)A(6,2): y−2=−35(x−6)y-2=\dfrac{-3}{5}(x-6)

  4. 5y−10=−3x+185y-10=-3x+18

  5. 3x+5y−28=03x+5y-28=0

Answer

3x+5y−28=03x+5y-28=0

Q10. Find the equation of a straight line which has slope −54\dfrac{-5}{4} and …

Find the equation of a straight line which has slope −54\dfrac{-5}{4} and passing through the point (−1,2)(-1,2).

  1. Point-slope form y−y1=m(x−x1)y-y_1=m(x-x_1) with m=−54m=\dfrac{-5}{4}, (−1,2)(-1,2):

  2. y−2=−54(x+1)y-2=\dfrac{-5}{4}(x+1)

  3. 4y−8=−5x−54y-8=-5x-5

  4. 5x+4y−3=05x+4y-3=0

Answer

5x+4y−3=05x+4y-3=0

Q11. You are downloading a song. The percent yy (in decimal form) of …

You are downloading a song. The percent yy (in decimal form) of mega bytes remaining to get downloaded in xx seconds is given by y=−0.1x+1y=-0.1x+1.

(i)

find the total MB of the song.

  1. At the start no part is downloaded, so put x=0x=0.

  2. y=−0.1(0)+1=1y=-0.1(0)+1=1

  3. y=1y=1 means the whole song, that is 11 MB, is remaining.

Answer

11 MB

(ii)

after how many seconds will 75%75\% of the song gets downloaded?

  1. If 75%75\% is downloaded, then 25%25\% remains, so y=0.25y=0.25.

  2. 0.25=−0.1x+10.25=-0.1x+1

  3. 0.1x=0.750.1x=0.75

  4. x=7.5x=7.5

Answer

7.57.5 seconds

(iii)

after how many seconds the song will be downloaded completely?

  1. Fully downloaded means nothing remains, so y=0y=0.

  2. 0=−0.1x+10=-0.1x+1

  3. 0.1x=10.1x=1

  4. x=10x=10

Answer

1010 seconds

Q12. Find the equation of a line whose intercepts on the xx and …

Find the equation of a line whose intercepts on the xx and yy axes are given below.

(i)

4, −64,\ -6

  1. Intercept form xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1 with a=4a=4, b=−6b=-6:

  2. x4+y−6=1\dfrac{x}{4}+\dfrac{y}{-6}=1

  3. Multiply by 1212: 3x−2y=123x-2y=12

  4. 3x−2y−12=03x-2y-12=0

Answer

3x−2y−12=03x-2y-12=0

(ii)

−5,34-5,\dfrac{3}{4}

  1. Intercept form with a=−5a=-5, b=34b=\dfrac34:

  2. x−5+y34=1\dfrac{x}{-5}+\dfrac{y}{\frac34}=1

  3. −x5+4y3=1\dfrac{-x}{5}+\dfrac{4y}{3}=1

  4. Multiply by 1515: −3x+20y=15-3x+20y=15

  5. 3x−20y+15=03x-20y+15=0

Answer

3x−20y+15=03x-20y+15=0

Q13. Find the intercepts made by the following lines on the coordinate axes.

(i)

3x−2y−6=03x-2y-6=0

  1. 3x−2y−6=0⇒3x−2y=63x-2y-6=0\Rightarrow 3x-2y=6

  2. Divide by 66: 3x6−2y6=1\dfrac{3x}{6}-\dfrac{2y}{6}=1

  3. x2+y−3=1\dfrac{x}{2}+\dfrac{y}{-3}=1

  4. Comparing with xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1.

Answer

xx intercept =2=2, yy intercept =−3=-3

(ii)

4x+3y+12=04x+3y+12=0

  1. 4x+3y+12=0⇒4x+3y=−124x+3y+12=0\Rightarrow 4x+3y=-12

  2. Divide by −12-12: 4x−12+3y−12=1\dfrac{4x}{-12}+\dfrac{3y}{-12}=1

  3. x−3+y−4=1\dfrac{x}{-3}+\dfrac{y}{-4}=1

  4. Comparing with xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1.

Answer

xx intercept =−3=-3, yy intercept =−4=-4

Q14. Find the equation of a straight line

(i)

passing through (1,−4)(1,-4) and has intercepts which are in the ratio 2:52:5

Show steps (7)

  1. Let the intercepts be a=2ka=2k and b=5kb=5k.

  2. x2k+y5k=1\dfrac{x}{2k}+\dfrac{y}{5k}=1

  3. 5x+2y=10k5x+2y=10k

  4. It passes through (1,−4)(1,-4): 5(1)+2(−4)=10k5(1)+2(-4)=10k

  5. −3=10k-3=10k

  6. So 5x+2y=−35x+2y=-3

  7. 5x+2y+3=05x+2y+3=0

Answer

5x+2y+3=05x+2y+3=0

(ii)

passing through (−8,4)(-8,4) and making equal intercepts on the coordinate axes

Show steps (6)

  1. Equal intercepts means a=ba=b.

  2. xa+ya=1⇒x+y=a\dfrac{x}{a}+\dfrac{y}{a}=1\Rightarrow x+y=a

  3. It passes through (−8,4)(-8,4): −8+4=a-8+4=a

  4. a=−4a=-4

  5. So x+y=−4x+y=-4

  6. x+y+4=0x+y+4=0

Answer

x+y+4=0x+y+4=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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