{"id":25415,"date":"2020-09-15T11:10:22","date_gmt":"2020-09-15T05:40:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25415"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-solutions-for-class-10-maths-chapter-7-ex-7-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-7-ex-7-4\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4<\/h2>\n

Question 1.
\nDetermine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
\nAnswer:
\nLet P(x1<\/sub>, y1<\/sub>) be common point of both lines and divide the line segment joining A(2, – 2) and B (3, 7) in ratio K : 1
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
\nAnswer:
\nGiven point are A(x, y), B(1, 2) & C(7, 0) These points will be collinear if the area of the triangle formed by them is zero.
\nArea of \u2206 ADC
\n\"KSEEB
\n0 = \\(\\frac{1}{2}\\)[x(2 – 0)+1(0 – y)+7(y – 2)]
\n0 = x \u00d7 2 + 1 \u00d7 – y + 7y – 14
\n0 = 2x – y + 7y – 14
\n2x + 6y – 14 = 0 divide by 2
\nx + 3y – 7 = 0
\nIt is the relation between x & y<\/p>\n

Question 3.
\nFind the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
\nAnswer:
\nLet A \u2192 (6,- 6), B \u2192 (3,- 7) and C \u2192 (3, 3)
\n\"KSEEB
\nSquaring
\n(x – 6)2<\/sup> + (y + 6)2<\/sup> = (x – 3)2<\/sup> + (y + 7)2<\/sup>
\nx2<\/sup> – 12x + 36 +y2<\/sup> +12y + 36
\n= x2<\/sup> – 6x + 9 + y2<\/sup> +14y + 49
\nx2<\/sup> – x2<\/sup> + y2<\/sup> – y2<\/sup> – 12x +6x + 12y – 14y + 72 – 58 = 0
\n– 6x – 2y = – 14 devide by – 2
\n3x + y = 7 \u2192 (1)
\n(x – 3)2<\/sup> +(y + 7)2<\/sup> = (x – 3)2<\/sup> +(y – 3)2<\/sup> (y + 7)2<\/sup> = (y – 3)2<\/sup>
\ny2<\/sup> +49 + 14y = y2<\/sup> + 9 – 6y
\ny2<\/sup> – y2<\/sup> + 14y + 6y = 9 – 49
\n20y = – 40
\ny = – 2
\nPutting y = – 2 in eqn (1) 3x + y = 7
\n3x – 2 = 7
\n3x = 7 + 2
\n3x = 9
\nx = \\(\\frac{9}{3}\\) = 3
\nx = 3
\nThus I(x, y) \u2192 (3, – 2)
\nHence, the centre of a circle is (3, – 2)<\/p>\n

Question 4.
\nThe two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
\nAnswer:
\nLet ABCD is a square where two opposite vertices are A(- 1, 2) & C(3,2)
\n\"KSEEB
\nAB = BC = CD = AD [ABCD is a square]
\nAB = BC
\n\"KSEEB
\nSquaring both side
\n(x +1 )2<\/sup> + (y – 2)2<\/sup> = (3 – x )2<\/sup> + (2 – y )2<\/sup>
\nx2<\/sup> + 2x + 1 + y2<\/sup> – 4y + 4
\n= 9 + x2<\/sup> – 6x + 4 + y2<\/sup> – 4y
\n2x + 6x – 4y + 4y = 13 – 5
\n8x = 8
\nx = \\(\\frac{8}{8}\\) = 1
\nx = 1
\nIn \u2206 ABC [B_ =90\u00b0
\n\"KSEEB
\nx2<\/sup> + 1 + 2x + y2<\/sup> + 4 – 4x + 9 + x2<\/sup>
\n– 6x + 4 + y2<\/sup> – 4y = (4)2<\/sup>
\n2x2<\/sup> – 4x + 2y2<\/sup> – 8y + 18 = 16
\n2x(1)2<\/sup> – 4(1) + 2y2<\/sup> – 8y + 18 – 16 = 0
\n2 – 4 + 2y2<\/sup> – 8y + 2 = 0
\n2y2<\/sup> – 8y = 0
\n2y(y – 4) = 0
\n2y = 0 & y – 4 = 0
\ny = 0 & y = 4
\nHence, the other two vertices are (x, y) &(x1<\/sub>, y1<\/sub>) are (1, 0) & (1, 4)<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nThe Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the, boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Fig.7.4 The students are to sow seeds of flowering plants on the remaining area of the plot.
\n\"KSEEB
\n(i) Taking A as origin,find the coordinates of the vertices of the triangle.
\n(ii) What will be the coordinates of the vertices of \u2206 PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?
\nAnswer:
\n(i) Taking A as origin then AD is x – axis and AB is y – axis.
\nCo-ordinates of P, Q and R are
\nP \u2192 (4, 6), Q \u2192 (3, 2) & R \u2192 (6, 5)
\narea of \u2206 PQR
\n\"KSEEB
\narea of \u2206 PQR
\n= \\(\\frac{1}{2}\\)[4(2 – 5) + 3(5 – 6) + (6 – 2)]
\n= \\(\\frac{1}{2}\\)[4 \u00d7 – 3 + 3 \u00d7 – 1 + 6 \u00d7 4]
\n= \\(\\frac{1}{2}\\)[- 12 – 3 + 24]
\narea of \u2206 PQR
\n= \\(\\frac{1}{2}\\)[- 15 + 24] = 9\/2 sq units<\/p>\n

(ii) Taking C as origin then CB is x – axis & CD is y – axis.
\nP(x1<\/sub>, y1<\/sub>) = (12, 2), Q(X2<\/sub>, y2<\/sub>) = (13, 6), and R(x3<\/sub>, y3<\/sub>) = (10, 3),
\nx1<\/sub> = 12, y1<\/sub> = 2, x2<\/sub> = 13, y2<\/sub> = 6, x3<\/sub> = 10 & y3<\/sub> = 3
\narea of \u2206 PQR
\n= \\(\\frac{1}{2}\\)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
\n= \\(\\frac{1}{2}\\)[12 x 3 + 13 x 1 + 10 x – 4]
\n= \\(\\frac{1}{2}\\)[36 + 13 – 40]
\n= \\(\\frac{1}{2}\\)[49 – 40] = 9\/2 Sq units
\nHence we observed that area of \u2206 remains same in both case.<\/p>\n

Question 6.
\nThe vertices of a \u2206 ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD\/AB = AE\/ AC = 1\/4 Calculate the area of the \u2206 ADE and compare it with the area of \u2206 ABC. (Recall Theorem 6.2 and Theorem 6.6).
\nAnswer:
\n\"KSEEB
\nD and E divide AB and AC respectively in the ration 1 : 3
\nSection Formula
\n\"KSEEB
\n\"KSEEB
\narea of \u2206 ABC
\n= \\(\\frac{1}{2}\\)[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]
\n= \\(\\frac{1}{2}\\)[4 \u00d7 3 + 1 \u00d7 – 4 + 7 \u00d7 1]
\n= \\(\\frac{1}{2}\\)[12 – 4 + 7]
\n= \\(\\frac{1}{2}\\) \u00d7 [19 – 4]
\n= \\(\\frac{15}{2}\\)sq units
\n\"KSEEB
\narea of ADE : area of ABC =1:16<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nLet A (4,2), B(6, 5) and C(1, 4) be the vertices of \u2206 ABC.
\n(i) The median from A meets BC at D. Find the coordinates of the point D.
\n(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
\n(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2:1.
\n(iv) What do yo observe?
\n[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
\n(v) If A(x1<\/sub>, y1<\/sub>), B(x2<\/sub>, y2<\/sub>) and C(x3<\/sub>, y3<\/sub>) are the vertices of \u2206 ABC, find the coordinates of the centroid of the triangle.
\nAnswer:
\n(i)
\n\"KSEEB
\nmedian AD of the triangle ABC divide the side BC into two equal parts.
\nTherefore D is the mid – point of side BC Co-ordinates of mid Point
\n\"KSEEB<\/p>\n

(ii) Form equation AP : PD = 2 : 1
\n\"KSEEB
\nSection Formula
\n\"KSEEB<\/p>\n

(iii) BQ : QE = 2 : 1
\n\"KSEEB
\nco-ordinates of
\n\\(\\frac{A D}{A B}=\\frac{A E}{E C}=\\frac{1}{4}\\)<\/p>\n

(iv) The co-ordinates P, Q, & R are same (11\/3, 11\/3) All these points represents the same point which is called centriod.<\/p>\n

(v) Point 0 is the centroid and AD is the midian. D is the mid point of BC and point 0 divide the AD into 2 : 1
\n\"KSEEB<\/p>\n

Question 8.
\nABCD is a rectangle formed by the points A(- 1, – 1), B(- 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
\nAnswer:
\nGiven: A \u2192 (- 1, – 1), B \u2192 (- 1, 4), C \u2192 (5, 4) & D \u2192 (5, – 1) Co-ordinates of midpoint
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB
\n\u2234 PQ = QR = RS = SP
\n\u2234 PQRS is either square (or) Rhomrus
\nDiagonal PR
\n\"KSEEB
\nand Diagonal QS
\n\"KSEEB
\nPR \u2260 QS
\n\u2234 Diagonal are not equal
\nTherefore, PQRS is a rhombus.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Question 1. Determine the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25415"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25415"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25415\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25415"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25415"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25415"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}