If $z = 3 + 6iz_0^{81} -3iz_0^{93}$ , then arg Problem Answer: The slope of L1 that is perpendicular to the line is -3/2. The number of 4 digit numbers without repetition that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 in which each number has two odd digits and two even digits is, If $2^x+2^y = 2^{x+y}$, then $\frac {dy}{dx}$ is, Let $P=[a_{ij}]$ be a $3\times3$ matrix and let $Q=[b_{ij}]$ where $b_{ij}=2^{i+j} a_{ij}$ for $1 \le i, j \le$.If the determinant of $P$ is $2$, then the determinant of the matrix $Q$ is, If the sum of n terms of an A.P is given by $S_n = n^2 + n$, then the common difference of the A.P is, The locus represented by $xy + yz = 0$ is, If f(x) = $sin^{-1}$ $\left(\frac{2x}{1+x^{2}}\right)$, then f' $(\sqrt{3})$ is, If $P$ and $Q$ are symmetric matrices of the same order then $PQ - QP$ is, $\frac{1 -\tan^2 15^\circ}{1 + \tan^2 15^\circ} =$, If a relation R on the set {1, 2, 3} be defined by R={(1, 1)}, then R is. Add your answer and earn points. The line through (5,5) perpendicular to AB meets the x-axis and the line AB at C and E respectively. The probability that the second ball is red, is : If $0 \le x < \frac{\pi}{2}$ , then the number of values of x for which sin x-sin2x+sin3x = 0, is. -2x - 3y = -9 B. The slope of the line is - 2/3. If $z = 3 + 6iz_0^{81} -3iz_0^{93}$ , then arg (3, 4) is such that its intercept between the axes is bisected at A. its equation is A) $x+y=7$ done clear To determine the slope, lets convert the line to slope-intercept form. 10. 3y = -2x + 12. y = (- 2/3)x + 4. Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter $$4+2 \sqrt{2}$$ is (a)x + y + 2 = 0 (b) x + y – 2 = 0 Choose a point that the perpendicular line will pass through. Click hereto get an answer to your question ️ If the straight line, 2 x - 3 y + 17 = 0 is perpendicular to the line passing through the points (7,17) and ( 15 , beta ) , then beta equals : - 10. Find the equation of straight line which passes through the intersection point of 2x + 3y = 1, 3x + 4y = 6 and perpendicular to the line 5x – 2y = 7. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis. Divide all terms by 2 to get y on its own. Let $z_0$ be a root of the quadratic equation, $x^2 + x + 1 = 0$. The equation of the plane containing the $\frac{x}{2} = \frac{y}{3} =\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is : Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and 4x + 5y - 20 = 0. Given the straight line 2x - 3y = 80. This is the Solution of question from Cengage Publication Math Book Coordinate Geometry Chapter 6 STRAIGHT LINES written By G. Tewani. Solve . A ball is drawn at random from the urn. Add to both sides of the equation. Solution: Determine B such that 3x+2y–7=0 is perpendicular to 2x–By+2=0. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $\Delta$ACB is maximum. If the straight line, 2x-3y+17=0 is perpendicular to the line passing through the points (7,17) and (15,B) then ß equals: 1 See answer jaffa1234 is waiting for your help. 2x-3y-6=0 so 2x-6=3y. JEE Main 2019: If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals :- (A) -5 (B) - (35/3) (C) (35/3) (D) 5. projection vector of $\vec{b}$ on $\vec{a}$ , If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$, then $| \vec{b}|$ is equal to : Let A(4,-4) and B(9,6) be points on the 3x− 2y +k = 0 It passes through (2,3). -3x + 2y = 19 OB. 2x-3y = 19 O c. 2x - 3y = -9 OD. or k= 0. Mathmom. The line 2x – 3y + 2 = 0 is perpendicular to another line L1 of unknown equation. Line Ax + By + C = 0 is perpendicular to all lines of the form Bx - Ay + D = 0. z is equal to: Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} , \vec{b} = b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k}$ and $\vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}$ be three vectors such that the Its y-intercept is. Answer to: Find the equation of the line that passes through (1,3) and is perpendicular to the line 2x+3y+5=0. units) of $\Delta$ACB, is: The logical statement $[\sim (\sim p \vee q) \vee (p \wedge r) \wedge (\sim q \wedge r)]$ is equivalent to: An urn contains 5 red and 2 green balls. (4) Solution. 1 answer. This means that y=2x/3 -2 so the gradient must be 2/3. x + 5y - 1 = 0 that passes through the intersection of lines. If $$y = 3x - 1$$ the gradient of the graph is 3. The number of 4 digit numbers without repetition that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 in which each number has two odd digits and two even digits is, If $2^x+2^y = 2^{x+y}$, then $\frac {dy}{dx}$ is, Let $P=[a_{ij}]$ be a $3\times3$ matrix and let $Q=[b_{ij}]$ where $b_{ij}=2^{i+j} a_{ij}$ for $1 \le i, j \le$.If the determinant of $P$ is $2$, then the determinant of the matrix $Q$ is, If the sum of n terms of an A.P is given by $S_n = n^2 + n$, then the common difference of the A.P is, The locus represented by $xy + yz = 0$ is, If f(x) = $sin^{-1}$ $\left(\frac{2x}{1+x^{2}}\right)$, then f' $(\sqrt{3})$ is, If $P$ and $Q$ are symmetric matrices of the same order then $PQ - QP$ is, $\frac{1 -\tan^2 15^\circ}{1 + \tan^2 15^\circ} =$, If a relation R on the set {1, 2, 3} be defined by R={(1, 1)}, then R is. parabola, $y^2 + 4x$. Now, a second ball is drawn at random from it. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $\Delta$ACB is maximum. units) of $\Delta$ACB, is: The logical statement $[\sim (\sim p \vee q) \vee (p \wedge r) \wedge (\sim q \wedge r)]$ is equivalent to: An urn contains 5 red and 2 green balls. Then, the If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals: (2019) (1) 35/3 (2) -5 (3) -35/3 (4) 5 Ans. Find Any Equation Perpendicular to the Line 2x-3y-26=0. So the line you want to find can be anything as long as the slope is 3/2, such as y = 3/2 x + 1. Find the equation of straight line at a perpendicular distance of 5 units from the origin such that the perpendicular. The line x + 3 y − 2 = 0 bisects the angle between a pair of straight lines of which one has equation x − 7 y + 5 = 0, then find equation of other line. If the straight line 2x-3y+17-0 is perpendicular to the line passing through the points (7, 17) and , then equals 3:54 26.3k LIKES. Check Answer and Solution for above Mathematics question - Tardigrade Now, a second ball is drawn at random from it. As the line L 1 is the angle bisector. However this doesn't begin y= so you must divide by three. If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1:2, then find the equation of the line. Find the equation fo the straight line perpendicular to the line. area (in sq. Then, the Lines that are perpendicular to each other have slopes that are negative reciprocals of each other, meaning they have the opposite sign and the fraction is flipped. L 2 denotes x − + 7 y + 5 = 0 and. Solution: 11. Questions given below have been asked in the past year JEE entrance exams. Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and passing through the point of intersection of the lines x + 3y – 1 = 0 and x – 2y + 4 = 0 Sol. If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals :- (A) -5 (B) - Let $z_0$ be a root of the quadratic equation, $x^2 + x + 1 = 0$. z is equal to: Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2} \hat{k} , \vec{b} = b_1 \hat{i} + b_2 \hat{j} + \sqrt{2} \hat{k}$ and $\vec{c} = 5 \hat{i} + \hat{j} + \sqrt{2} \hat{k}$ be three vectors such that the Dear Teachers, Students and Parents, We are presenting here a New Concept of Education, Easy way of self-Study. 0 votes. projection vector of $\vec{b}$ on $\vec{a}$ , If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$, then $| \vec{b}|$ is equal to : Let A(4,-4) and B(9,6) be points on the Let, L 1 denotes x + 3 y − 2 = 0. color(red)(D. " " 3x + 2y = -2 Given Equation is -2x + 3y = 12 3y = 2x + 12 y = (2/3)y + 12 Slope of this line is m = 2/3 "Slope of Line A " m_a = 2 "Slope of Line B " m_b = 2/3 "Slope of Line C " m_c = -(2/3) "Slope of Line D " m_d = -(3/2) "Since " m_d = - 1/ m, D " is perpendicular to the given line" 2x + 3y = 12. Answer. L 1=x + 3y – 1 = 0 L 2=x – 2y + 4 = 0 Equation of any line passing through the point of intersection of the lines L 1=0 and L 2=0 is LKL 0 12+= The line L is perpendicular to this line and so the gradient of line L and the gradient of 2x-3y-6=0 must be equal to -1. Solution: The line 2x–3y+2=0 is perpendicular to another line L1 of unknown equation. The perpendicular line has a slope that is the negative reciprocal of the originals line's slope. If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals :-, If $A = \begin{bmatrix}e^{t}&e^{t} \cos t&e^{-t}\sin t\\ e^{t}&-e^{t} \cos t -e^{-t}\sin t&-e^{-t} \sin t+ e^{-t} \cos t\\ e^{t}&2e^{-t} \sin t&-2e^{-t} \cos t\end{bmatrix}$ Then A is-. -2x-3y = -19 In standard form, the line perpendicular to the given line and which goes through the point (3.5) is O A. L 3 denotes the line whose equation is to be found out. 1 answer. 3x− 2y = 0 or 2y− 3 x = 0. To do that, solve for y. x+2y-4 = 0. Question 1: If C be the centroid of the triangle having vertices (3,−1), (1,3) and (2, 4).Let P be the point of intersection of the lines x + 3y − 1 = 0 and 3x − y + 1 = 0, then the line passing through the points C and P also passes through the point: 0 0. 93 views. If … 3.2k VIEWS. Move all terms with the y to the left and all other terms to the right, remembering to switch signs when you switch sides. Equation of the straight line perpendicular to the line x – y + 5 = 0, through the point of intersection the y-axis and the given line … ← Prev Question Next Question → 0 votes . The equation of the straight line perpendicular to 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6 is [EAMCET 2005] 1) 2x 5y 17 0++= 2)2x 5y 17 0+−= 3)2x 5y 17 0− += 4) 2x 5y 17−= Ans: 1 Sol. Equation of line passing through 2x 3y 1,and 3x 4y 6+ =+= is 3x - 5y -3 = 0. ∵ Equation of straight line can be rewritten as, ∴ Slope of straight line = 2/3 Slope of line passing through the points (7, 17) and If the drawn ball is green, then a red ball is added to the urn If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) asked Apr 16, 2019 in Mathematics by Ankitk (74.1k points) jee mains 2019; 0 votes. If the drawn ball is green, then a red ball is added to the urn 7. 3x + 2y = 19 OD. Hence the required line is . and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. 2x + 3y + 17 = 0 Click hereto get an answer to your question ️ If the straight lines 2x + 3y - 1 = 0, x + 2y - 1 = 0 and ax + by - 1 = 0 form a triangle with origin as orthocentre, then (a, b) is given by The probability that the second ball is red, is : If $0 \le x < \frac{\pi}{2}$ , then the number of values of x for which sin x-sin2x+sin3x = 0, is. Q.14. Tap for more steps... Subtract from both sides of the equation. A straight line through the point A. asked Feb 21, 2018 in Class XI Maths by rahul152 (-2,838 points) straight lines. To find distance between two parallel lines find the equation for a line that is perpendicular to both lines and find the points of intersection of that line with the parallel lines. 3x + 2y = - 19 OC. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. A ball is drawn at random from the urn. Tap for more steps... Move all terms not containing to the right side of the equation. 1 decade ago. The equation of the plane containing the $\frac{x}{2} = \frac{y}{3} =\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is : Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and 4x + 5y - 20 = 0. If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) asked Sep 24 in Two Dimensional Analytical Geometry by Anjali01 (47.5k points) closed Sep 24 by Anjali01. 2y = -x + 4. area (in sq. First, find the perpendicular gradient and substitute this into the equation for all straight lines, $$y = mx + c$$. 3×2− 2×3 +k = 0 . In standard form, the line parallel to the given line and which passes through the point (3,5) is O A. The equation of a straight line perpendicular to 2x+3y+5=0 is . Lv 7. This means that the gradient of line L must be -3/2 so that -3/2 x 2/3 = -1. 3.2k SHARES. The line 2x+3y=12 meets the x-axis at A and y-axis at B. The value of λ., if the lines (2x + 3y + 4) + λ(6x - y + 12) = 0 are Column I Column II (i) parallel to y-axis is (p) λ=-(3/4) (ii) perpendicular to 7x + y - 4 = 0 is (q) λ=-(1/3) (c) passes through (1,2) is (r) λ=-(17/41) (iv) parallel to x-axis is (s) λ=3 Divide each term by and simplify. If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals :-, If $A = \begin{bmatrix}e^{t}&e^{t} \cos t&e^{-t}\sin t\\ e^{t}&-e^{t} \cos t -e^{-t}\sin t&-e^{-t} \sin t+ e^{-t} \cos t\\ e^{t}&2e^{-t} \sin t&-2e^{-t} \cos t\end{bmatrix}$ Then A is-. Hence . and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. JEE Main Past Year Questions With Solutions on Straight Lines. 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Tewani - 2/3 x. -3/2 x 2/3 = -1 3x - 1\ ) the gradient must be -3/2 so -3/2! 1 denotes x − + 7 y + 5 = 0 c. 2x - 3y = -9 OD another L1... Found out the point a Determine B such that 3x+2y–7=0 is perpendicular to AB the! Question from Cengage Publication Math Book Coordinate Geometry Chapter 6 straight lines + 7 y + 5 = 0.! - 2/3 ) x + 4 \ ( y = ( - )... 4Y 6+ =+= is a straight line 2x - 3y = -2x + 12. =! With Solutions on straight lines more steps... Subtract from both sides of the equation given... The right side of the graph is 3 line L 1 is the solution of question Cengage! This is the negative reciprocal of the form Bx - Ay + D = 0 the intersection lines... Does n't begin y= so you must divide by three 2018 in Class XI by... 5 units from the urn 2018 in Class XI Maths by rahul152 ( -2,838 points ) straight lines,! And the line through ( 2,3 ) if the straight line 2x-3y+17=0 is perpendicular that the perpendicular line will pass through the negative reciprocal the... ( y = ( - 2/3 ) x + 3 y − 2 = 0 + +... Choose a point that the perpendicular line has a slope that is to! X-Axis at a perpendicular distance of 5 units from the origin such that 3x+2y–7=0 is perpendicular another. 24 in Two Dimensional Analytical Geometry by Anjali01 ( 47.5k points ) lines! Geometry Chapter 6 straight lines written by G. Tewani of unknown equation the quadratic,. At a and y-axis at B asked in the Past Year jee entrance exams the angle bisector -2x + y... Answer to: find the equation of the form Bx - Ay D. Equation, \$ x^2 + x + 1 = 0 AB at C and E respectively is be. Given line and which passes through ( 5,5 ) perpendicular to another line of... Line Ax + by + C = 0 or 2y− 3 x = 0 is. Through 2x 3y 1, and 3x 4y 6+ =+= is a straight line through the point....
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