{"id":919,"date":"2023-10-31T07:02:49","date_gmt":"2023-10-31T01:32:49","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=919"},"modified":"2023-11-10T11:20:25","modified_gmt":"2023-11-10T05:50:25","slug":"samacheer-kalvi-10th-maths-guide-chapter-5-ex-5-4","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-10th-maths-guide-chapter-5-ex-5-4\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 5 Coordinate Geometry Ex 5.4"},"content":{"rendered":"

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4<\/h2>\n

Question 1.
\nFind the slope of the following straight lines.
\n(i) 5y – 3 = 0
\n(ii) 7x – \\(\\frac { 3 }{ 17 } \\) = 0
\nSolution:
\n(i) 5y – 3 = 0
\n5y = 3 \u21d2 y = \\(\\frac { 3 }{ 5 } \\)
\nSlope = 0<\/p>\n

(ii) 7x – \\(\\frac { 3 }{ 17 } \\) = 0 (Comparing with y = mx + c)
\n7x = \\(\\frac { 3 }{ 17 } \\)
\nSlope is undefined<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nFind the slope of the line which is
\n(i) parallel to y = 0.7x – 11
\n(ii) perpendicular to the line x = -11
\nSolution:
\n(i) y = 0.7x – 11
\nSlope = 0.7 (Comparing with y = mx + c)
\n(ii) Perpendicular to the line x = – 11
\nSlope is undefined (Since the line is intersecting the X-axis)<\/p>\n

Question 3.
\nCheck whether the given lines are parallel or perpendicular
\n(i) \\(\\frac { x }{ 3 } \\) + \\(\\frac { y }{ 4 } \\) + \\(\\frac { 1 }{ 7 } \\) = 0 and \\(\\frac { 2x }{ 3 } \\) + \\(\\frac { y }{ 2 } \\) + \\(\\frac { 1 }{ 10 } \\) = 0
\n(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
\nSolution:
\n(i) \\(\\frac { x }{ 3 } \\) + \\(\\frac { y }{ 4 } \\) + \\(\\frac { 1 }{ 7 } \\) = 0 ; \\(\\frac { 2x }{ 3 } \\) + \\(\\frac { y }{ 2 } \\) + \\(\\frac { 1 }{ 10 } \\) = 0
\nSlope of the line (m1<\/sub>) = \\(\\frac { -a }{ b } \\)
\n= – \\(\\frac { 1 }{ 3 } \\) \u00f7 \\(\\frac { 1 }{ 4 } \\) = –\\(\\frac { 1 }{ 3 } \\) \u00d7 \\(\\frac { 4 }{ 1 } \\) = – \\(\\frac { 4 }{ 3 } \\)
\nSlope of the line (m2<\/sub>) = – \\(\\frac { 2 }{ 3 } \\) \u00f7 \\(\\frac { 1 }{ 2 } \\) = –\\(\\frac { 2 }{ 3 } \\) \u00d7 \\(\\frac { 2 }{ 1 } \\) = – \\(\\frac { 4 }{ 3 } \\)
\nm1<\/sub> = m2<\/sub> = – \\(\\frac { 4 }{ 3 } \\)
\n\u2234 The two lines are parallel.<\/p>\n

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
\nSlope of the line (m1<\/sub>) = \\(\\frac { -5 }{ 23 } \\)
\nSlope of the line (m2<\/sub>) = \\(\\frac { -23 }{ -5 } \\) = \\(\\frac { 23 }{ 5 } \\)
\nm1<\/sub> \u00d7 m2<\/sub> = \\(\\frac { -5 }{ 23 } \\) \u00d7 \\(\\frac { 23 }{ 5 } \\) = -1
\n\u2234 The two lines are perpendicular<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nIf the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
\nSolution:
\nSlope of the first line 12y = -(p + 3)x +12
\ny = \\(-\\frac{(p+3) x}{12}+1\\) (Comparing with y = mx + c)
\nSlope of the second line (m1<\/sub>) = \\(\\frac { -(p+3) }{ 12 } \\)
\nSlope of the second line 12x – 7y = 16
\n(m2<\/sub>) = \\(\\frac { -a }{ b } \\) = \\(\\frac { -12 }{ -7 } \\) = \\(\\frac { 12 }{ 7 } \\)
\nSince the two lines are perpendicular
\nm1<\/sub> \u00d7 m2<\/sub> = -1
\n\\(\\frac { -(p+3) }{ 12 } \\) \u00d7 \\(\\frac { 12 }{ 7 } \\) = -1 \u21d2 \\(\\frac { -(p+3) }{ 7 } \\) = -1
\n-(p + 3) = -7
\n– p – 3 = -7 \u21d2 -p = -7 + 3
\n-p = -4 \u21d2 p = 4
\nThe value of p = 4<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nFind the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
\nSolution:
\nSlope of the line = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
\nSlope of the line QR = \\(\\frac { 4+2 }{ -5-3 } \\) = \\(\\frac { 6 }{ -8 } \\) = \\(\\frac { 3 }{ -4 } \\) \u21d2 – \\(\\frac { 3 }{ 4 } \\)
\nSlope of its parallel = – \\(\\frac { 3 }{ 4 } \\)
\nThe given point is p(-5, 2)
\nEquation of the line is y – y1<\/sub> = m(x – x1<\/sub>)
\ny – 2 = – \\(\\frac { 3 }{ 4 } \\) (x + 5)
\n4y – 8 = -3x – 15
\n3x + 4y – 8 + 15 = 0
\n3x + 4y + 7 = 0
\nThe equation of the line is 3x + 4y + 7 = 0<\/p>\n

Question 6.
\nFind the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
\nSolution:
\nLet the vertices A (6, 7), B (2, -3), D (6, -2)
\n\"Samacheer
\nSlope of a line = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
\nSlope of AB = \\(\\frac { -3-7 }{ 2-6 } \\) = \\(\\frac { -10 }{ -4 } \\) = \\(\\frac { 5 }{ 2 } \\)
\nSlope of its perpendicular (CD) = – \\(\\frac { 2 }{ 5 } \\)
\nEquation of the line CD is y – y1<\/sub> = m(x – x1<\/sub>)
\ny + 2 = –\\(\\frac { 2 }{ 5 } \\) (x – 6)
\n5(y + 2) = -2 (x – 6)
\n5y + 10 = -2x + 12
\n2x + 5y + 10 – 12 = 0
\n2x + 5y – 2 = 0
\nThe equation of the line is 2x + 5y – 2 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nA(-3,0) B(10, -2) and C(12,3) are the vertices of \u2206ABC. Find the equation of the altitude through A and B.
\nSolution:
\nTo find the equation of the altitude from A.
\nThe vertices of \u2206ABC are A(-3, 0), B(10, -2) and C(12, 3)
\n\"Samacheer
\nSlope of BC = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
\n= \\(\\frac { 3+2 }{ 12-10 } \\) = \\(\\frac { 5 }{ 2 } \\)
\nSlope of the altitude AD is – \\(\\frac { 2 }{ 5 } \\)
\nEquation of the altitude AD is
\ny – y1<\/sub> = m (x – x1<\/sub>)
\ny – 0 = – \\(\\frac { 2 }{ 5 } \\) (x + 3)
\n5y = -2x -6
\n2x + 5y + 6 = 0
\nEquation of the altitude AD is 2x + 5y + 6 = 0
\nEquation of the altitude from B
\n\"Samacheer
\nSlope of AC = \\(\\frac { 3-0 }{ 12+3 } \\) = \\(\\frac { 3 }{ 15 } \\) = \\(\\frac { 1 }{ 5 } \\)
\nSlope of the altitude AD is -5
\nEquation of the altitude BD is y – y1<\/sub>= m (x – x1<\/sub>)
\n7 + 2 = -5 (x – 10)
\ny + 2 = -5x + 50
\n5x + 7 + 2 – 50 = 0 \u21d2 5x + 7 – 48 = 0
\nEquation of the altitude from B is 5x + y – 48 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nFind the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
\nSolution:
\n“C” is the mid point of AB also CD \u22a5 AB.
\n\"Samacheer
\nSlope of AB = \\(\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\)
\n= \\(\\frac { -4-2 }{ 6+4 } \\) = \\(\\frac { -6 }{ 10 } \\) = – \\(\\frac { 3 }{ 5 } \\)
\nSlope of the \u22a5r<\/sup> AB is \\(\\frac { 5 }{ 3 } \\)
\nMid point of AB = (\\(\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}\\))
\n= (\\(\\frac { -4+6 }{ 2 } \\),\\(\\frac { 2-4 }{ 2 } \\)) = (\\(\\frac { 2 }{ 2 } \\),\\(\\frac { -2 }{ 2 } \\)) = (1,-1)
\nEquation of the perpendicular bisector of CD is
\ny – y1<\/sub> = m(x – x1<\/sub>)
\ny + 1 = \\(\\frac { 5 }{ 3 } \\) (x – 1)
\n5(x – 1) = 3(y + 1)
\n5x – 5 = 3y + 3
\n5x – 3y – 5 – 3 = 0
\n5x – 3y – 8 = 0
\nEquation of the perpendicular bisector is 5x – 3y – 8 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nFind the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
\nSolution:
\nGiven lines are.
\n\"Samacheer
\nx = \\(\\frac { 43 }{ 43 } \\) = 1
\nSubstitute the value of x = 1 in (1)
\n7(1) + 3y = 10 \u21d2 3y = 10 – 7
\ny = \\(\\frac { 3 }{ 3 } \\) = 1
\nThe point of intersection is (1,1)
\nEquation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
\nThis line passes through (1,1)
\n13 (1) + 5 (1) + k = 0
\n13 + 5 + k = 0 \u21d2 18 + k = 0
\nk = -18
\n\u2234 The equation of the line is 13x + 5y – 18 = 0<\/p>\n

Question 10.
\nFind the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
\nSolution:
\nGiven lines are.
\n\"Samacheer
\nSubstitute the value of x = \\(\\frac { 16 }{ 7 } \\) in (2)
\n3 \u00d7 \\(\\frac { 16 }{ 7 } \\) + 2y = 10 \u21d2 2y = 10 – \\(\\frac { 48 }{ 7 } \\)
\n2y = \\(\\frac { 70-48 }{ 7 } \\) \u21d2 2y = \\(\\frac { 22 }{ 7 } \\)
\ny = \\(\\frac{22}{2 \\times 7}\\) = \\(\\frac { 11 }{ 7 } \\)
\nThe point of intersect is (\\(\\frac { 16 }{ 7 } \\),\\(\\frac { 11 }{ 7 } \\))
\nEquation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
\nThis line passes through (\\(\\frac { 16 }{ 7 } \\),\\(\\frac { 11 }{ 7 } \\))
\n7 (\\(\\frac { 16 }{ 7 } \\)) + 4 (\\(\\frac { 11 }{ 7 } \\)) + k = 0 \u21d2 16 + \\(\\frac { 44 }{ 7 } \\) + k = 0
\n\\(\\frac { 112+44 }{ 7 } \\) + k = 0 \u21d2 \\(\\frac { 156 }{ 7 } \\) + k = 0
\nk = – \\(\\frac { 156 }{ 7 } \\)
\nEquation of the line is 7x + 4y – \\(\\frac { 156 }{ 7 } \\) = 0
\n49x + 28y – 156 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nFind the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
\nSolution:
\nThe given lines are.
\n\"Samacheer
\n\"Samacheer
\nSubstitute the value of x = 0 in (1)
\n3 (0) + y = -2
\ny = -2
\nThe point of intersection is (0, -2).
\nThe given equation is
\n\"Samacheer
\nSubstitute the value of y = \\(\\frac { 9 }{ 11 } \\) in (6)
\n– x + 2 (\\(\\frac { 9 }{ 11 } \\)) = 3 \u21d2 -x + \\(\\frac { 18 }{ 11 } \\) = 3
\n-x = 3 – \\(\\frac { 18 }{ 11 } \\) = \\(\\frac { 33-18 }{ 11 } \\) = \\(\\frac { 15 }{ 11 } \\)
\nx = – \\(\\frac { 15 }{ 11 } \\)
\nThe point of intersection is (-\\(\\frac { 15 }{ 11 } \\),\\(\\frac { 9 }{ 11 } \\))
\nEquation of the line joining the points (0, -2) and (-\\(\\frac { 15 }{ 11 } \\),\\(\\frac { 9 }{ 11 } \\)) is
\n\"Samacheer
\n\"Samacheer
\n31 \u00d7 (- 11x) = 11 \u00d7 15 (y + 2) = 165 (y + 2)
\n– 341 x = 165 y + 330
\n– 341 x – 165 y – 330 = 0
\n341 x + 165 y + 330 = 0
\n(\u00f7 by 11) \u21d2 31 x + 15 y + 30 = 0
\nThe required equation is 31 x + 15 y + 30 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nFind the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
\nSolution:
\nGiven lines are.
\n8x + 3y = 18 …..(1)
\n4x + 5y = 9 …..(2)
\n\"Samacheer
\nx = \\(\\frac { 63 }{ 28 } \\) = \\(\\frac { 9 }{ 4 } \\)
\nSubstitute the value of x = \\(\\frac { 9 }{ 4 } \\) in (2)
\n4 (\\(\\frac { 9 }{ 4 } \\)) + 5y = 9
\n9 + 5y = 9 \u21d2 5y = 9 – 9
\n5y = 0 \u21d2 y = 0
\nThe point of intersection is (\\(\\frac { 9 }{ 4 } \\),0)
\nMid point of the points (5, -4) and (-7, 6)
\n\"Samacheer
\nEquation of the line joining the points (\\(\\frac { 9 }{ 4 } \\),0) and (-1,1)
\n\"Samacheer
\n-13y = 4x – 9
\n-4x – 13y + 9 = 0 \u21d2 4x + 13y – 9 = 0
\nThe equation of the line is 4x + 13y – 9 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/919"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=919"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/919\/revisions"}],"predecessor-version":[{"id":49381,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/919\/revisions\/49381"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=919"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=919"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}