{"id":6910,"date":"2021-10-21T22:29:15","date_gmt":"2021-10-21T16:59:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6910"},"modified":"2021-10-25T09:45:45","modified_gmt":"2021-10-25T04:15:45","slug":"let-a-b-c-and-d-be-non-zero-numbers-if-the-point-of-intersection-of-the-lines-4ax2ayc0-and-5bx2byd0-lies-in-the-fourth-quadrant-and-is-equidistant-from-the-two-axes-then","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/let-a-b-c-and-d-be-non-zero-numbers-if-the-point-of-intersection-of-the-lines-4ax2ayc0-and-5bx2byd0-lies-in-the-fourth-quadrant-and-is-equidistant-from-the-two-axes-then\/","title":{"rendered":"Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then"},"content":{"rendered":"

Solution :<\/h2>\n

Since, (\\(\\alpha\\), -\\(\\alpha\\)) lie on 4ax+2ay+c=0 and 5bx+2by+d=0.<\/p>\n

\\(\\therefore\\) 4a\\(\\alpha\\) + 2a\\(\\alpha\\) + c = 0 \\(\\implies\\) \\(\\alpha\\) = \\(-c\\over 2a\\)\u00a0 …..(i)<\/p>\n

Also, 5b\\(\\alpha\\) – 2b\\(\\alpha\\) + d = 0 \\(\\implies\\) \\(\\alpha\\) = \\(-d\\over 3b\\)\u00a0 \u00a0 …..(i)<\/p>\n

from equation (i) and (ii),<\/p>\n

\\(-c\\over 2a\\) = \\(-d\\over 3b\\)<\/p>\n

3bc = 2ad<\/p>\n

\n

Similar Questions<\/h3>\n

Find the distance between the line 12x \u2013 5y + 9 = 0 and the point (2,1)<\/a><\/p>\n

If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3:2, then k is equal to<\/a><\/p>\n

The x-coordinate of the incenter of the triangle that has the coordinates of mid-point of its sides as (0,1), (1,1) and (1,0) is<\/a><\/p>\n

If p is the length of the perpendicular from the origin to the line \\(x\\over a\\) + \\(y\\over b\\) = 1, then prove that \\(1\\over p^2\\) = \\(1\\over a^2\\) + \\(1\\over b^2\\)<\/a><\/p>\n

If PS is the median of the triangle, with vertices of P(2,2), Q(6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Since, (\\(\\alpha\\), -\\(\\alpha\\)) lie on 4ax+2ay+c=0 and 5bx+2by+d=0. \\(\\therefore\\) 4a\\(\\alpha\\) + 2a\\(\\alpha\\) + c = 0 \\(\\implies\\) \\(\\alpha\\) = \\(-c\\over 2a\\)\u00a0 …..(i) Also, 5b\\(\\alpha\\) – 2b\\(\\alpha\\) + d = 0 \\(\\implies\\) \\(\\alpha\\) = \\(-d\\over 3b\\)\u00a0 \u00a0 …..(i) from equation (i) and (ii), \\(-c\\over 2a\\) = \\(-d\\over 3b\\) 3bc = 2ad Similar Questions Find …<\/p>\n

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,48],"tags":[],"yoast_head":"\nLet a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/let-a-b-c-and-d-be-non-zero-numbers-if-the-point-of-intersection-of-the-lines-4ax2ayc0-and-5bx2byd0-lies-in-the-fourth-quadrant-and-is-equidistant-from-the-two-axes-then\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then\" \/>\n<meta property=\"og:description\" content=\"Solution : Since, ((alpha), -(alpha)) lie on 4ax+2ay+c=0 and 5bx+2by+d=0. (therefore) 4a(alpha) + 2a(alpha) + c = 0 (implies) (alpha) = (-cover 2a)\u00a0 …..(i) Also, 5b(alpha) – 2b(alpha) + d = 0 (implies) (alpha) = (-dover 3b)\u00a0 \u00a0 …..(i) from equation (i) and (ii), (-cover 2a) = (-dover 3b) 3bc = 2ad Similar Questions Find … Let a, b, c and d be non-zero numbers. 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